Explanation:
Okay, so Paul and Jesse each choose a number at random from the first six primes. I need to find the probability that the sum of the numbers they choose is even. Hmm, let me think about how to approach this.

First, let me recall the first six prime numbers. The primes start at 2, right? So the first six primes are 2, 3, 5, 7, 11, and 13. Let me write those down:

First six primes: 2, 3, 5, 7, 11, 13.

Okay, so both Paul and Jesse are choosing one of these numbers at random. Assuming that each number is equally likely to be chosen, and their choices are independent, right? So each has 6 possible choices, so there are 6 x 6 = 36 total possible pairs of numbers they could choose.

Now, the question is about the probability that their sum is even. To find this probability, I need to figure out how many of these 36 pairs result in an even sum, and then divide that by 36.

So, probability = (number of favorable outcomes) / (total number of outcomes) = (number of pairs with even sum) / 36.

Alright, so how can the sum of two numbers be even? Well, I remember that even + even = even, odd + odd = even, and even + odd = odd. So, for the sum to be even, both numbers have to be even or both have to be odd.

Therefore, the problem reduces to finding the number of pairs where both numbers are even or both are odd. Then, adding those two counts together, and dividing by 36.

So, let me figure out how many even numbers and how many odd numbers are in the first six primes.

Looking at the list: 2, 3, 5, 7, 11, 13.

Let me check each number's parity:

- 2 is even.
- 3 is odd.
- 5 is odd.
- 7 is odd.
- 11 is odd.
- 13 is odd.

So, there's only one even prime number, which is 2. All the others are odd. Therefore, in the first six primes, there is 1 even number and 5 odd numbers.

Therefore, the number of ways both Paul and Jesse can pick even numbers: since there's only 1 even number, Paul can choose 2, and Jesse can also choose 2. So, how many such pairs? Well, since Paul has 1 choice and Jesse has 1 choice, so 1 x 1 = 1 pair where both numbers are even.

Wait, but hold on. Wait, if they are choosing numbers at random, does that mean that they can choose the same number? For example, can both Paul and Jesse choose 2? I think so, since there's no restriction mentioned. The problem says they choose a number at random from the first six primes. So repetition is allowed? Or is it without replacement? Wait, the problem doesn't specify that they can't choose the same number. So, I think it's with replacement. Therefore, the total number of possible pairs is indeed 6 x 6 = 36, and each number is equally likely for both Paul and Jesse.

Therefore, the number of pairs where both numbers are even is 1 (for Paul) times 1 (for Jesse) = 1 pair.

Now, the number of pairs where both numbers are odd. There are 5 odd numbers. So, Paul can choose any of the 5 odd numbers, and Jesse can also choose any of the 5 odd numbers. So, that's 5 x 5 = 25 pairs.

Therefore, total number of favorable outcomes is 1 + 25 = 26.

Therefore, the probability is 26/36, which simplifies to 13/18.

Wait, but let me check again. Let me think about whether that's correct.

Alternatively, maybe I can approach this problem by thinking about the probability that both numbers are even or both are odd.

So, probability that both are even: probability Paul picks even times probability Jesse picks even. Since there's 1 even out of 6, so (1/6)*(1/6) = 1/36.

Probability that both are odd: probability Paul picks odd times probability Jesse picks odd. There are 5 odd numbers, so (5/6)*(5/6) = 25/36.

Adding those two probabilities: 1/36 + 25/36 = 26/36 = 13/18. So, same result. Therefore, the probability is 13/18.

Hmm, that seems straightforward, but let me check if there's a possibility that I made an error in counting the number of even and odd primes.

Wait, primes are numbers greater than 1 that have no divisors other than 1 and themselves. The first prime is 2, which is even, and all other primes are odd because if they were even, they would be divisible by 2. So, indeed, except for 2, all primes are odd. Therefore, in the first six primes, only 2 is even, so 1 even, 5 odd. That seems correct.

Another way to think about it: The sum is even if both numbers are even or both are odd. Since there is only one even number, the number of ways both are even is 1*1=1. The number of ways both are odd is 5*5=25. So, total favorable is 26. Total possibilities 36. 26/36=13/18≈0.722...

Alternatively, maybe I can think of it as the probability that the sum is even is equal to the probability that both numbers are even plus the probability that both are odd. Which is exactly what I did.

Alternatively, maybe think in terms of parity. Let me think.

Let me note that the parity of the sum depends on the parities of the numbers. So, if I think of each person's number as either even or odd.

So, Paul's number can be even with probability 1/6, odd with probability 5/6. Same for Jesse.

Then, the probability that both are even is (1/6)(1/6) = 1/36, and both odd is (5/6)(5/6)=25/36. So, adding those gives the same 26/36.

Alternatively, maybe think of it as the sum is even if they have the same parity. Therefore, the probability that they have the same parity.

Alternatively, maybe compute the probability that they have different parities, and subtract from 1.

So, probability that sum is even = 1 - probability that sum is odd.

Probability that sum is odd is probability that one is even and the other is odd.

Which would be (probability Paul even and Jesse odd) + (probability Paul odd and Jesse even).

Which is (1/6 * 5/6) + (5/6 * 1/6) = 5/36 + 5/36 = 10/36.

Therefore, probability sum is even is 1 - 10/36 = 26/36 = 13/18.

Same result. So, that cross-check works.

Therefore, seems like 13/18 is the answer.

But let me think again, just to make sure. Maybe list out all possibilities? Although 36 possibilities is a bit tedious, but maybe check for some cases.

Wait, if there's only one even number, 2. So, the possible pairs where both are even: only (2,2). So, that's one pair.

Pairs where both are odd: there are 5 odd numbers, so 5x5=25 pairs. So, 25 pairs where both are odd.

Therefore, total of 26 pairs.

Total possible pairs: 36. So, 26/36 reduces to 13/18.

Alternatively, maybe there's a mistake here. Let me think: is there any possibility that when they pick numbers, they can't pick the same number? Wait, the problem says "choose a number at random from the first six primes." It doesn't specify whether they are choosing with replacement or without replacement.

Wait, but usually, when problems say "choose a number at random" without specifying, it's usually with replacement. But sometimes, it can be interpreted as choosing two different numbers. However, in this case, since it's about the sum, and the problem doesn't mention anything about distinct numbers, it's safer to assume that they can choose the same number. For example, if they were picking two different numbers, the total number of possibilities would be 6*5=30. However, since the problem didn't specify distinctness, and given that in most probability problems unless stated otherwise, selections are with replacement.

But wait, actually, if they are choosing numbers independently, then it's with replacement. So, for example, if Paul chooses a number and then Jesse chooses a number, they can choose the same number, because they are independent choices. Therefore, the total number of possibilities is 6*6=36.

Therefore, the initial reasoning holds. So, 26/36 is correct.

But just to be thorough, let me check if maybe the problem is from a source where "choosing a number at random" implies distinct numbers. But since it's not specified, I think with replacement is the standard assumption.

Alternatively, if we assume that they are choosing two distinct numbers, then the total number of possible outcomes would be 6*5=30. Then, how many pairs would be both even or both odd?

But in that case, since there is only one even number, there's no way to have two even numbers if they have to be distinct. Therefore, the number of pairs where both are even would be zero. Then, the number of pairs where both are odd would be C(5,2) = 10, but since order matters (since Paul and Jesse are distinct), it's 5*4=20. Therefore, total favorable outcomes would be 20, and total is 30. Therefore, probability would be 20/30 = 2/3 ≈ 0.666..., which is different from 13/18≈0.722.

But since the problem says "each choose a number at random from the first six primes", and doesn't mention anything about distinct numbers, it's more likely that they can choose the same number. Therefore, the answer is 13/18.

But just to make sure, let me check if there's any other possible mistake. For example, maybe the first six primes are different? Wait, first six primes are definitely 2, 3, 5, 7, 11, 13. Let me check that.

Primes are numbers greater than 1 that have no positive divisors other than 1 and themselves.

First prime: 2

Second: 3

Third: 5

Fourth: 7

Fifth: 11

Sixth: 13. Yeah, correct.

So, only 2 is even, rest are odd. Therefore, as before.

Another possible mistake: thinking that the sum is even requires both numbers even, but actually odd + odd is also even. So, that's correct.

So, with that, the number of favorable cases is 1 + 25 = 26.

Another check: 26 divided by 36. Simplify by dividing numerator and denominator by 2: 13/18.

Yes.

Alternatively, if I think in terms of probability, as I did before, 1/36 + 25/36 = 26/36 = 13/18.

Therefore, the answer should be 13/18.

Therefore, I think that's the correct answer.

**Final Answer**
The probability that the sum of the numbers they choose is even is \boxed{\dfrac{13}{18}}.

To determine the probability that the sum of the numbers chosen by Paul and Jesse is even, we begin by identifying all the numbers involved and analyzing the relevant properties of even and odd sums.

---

### Step 1: Identify the Numbers

The first six prime numbers are:
$$
\{2, 3, 5, 7, 11, 13\}
$$
Among these:
- **Even number:** 2 (only one is even)
- **Odd numbers:** 3, 5, 7, 11, 13 (five in total)

---

### Step 2: Understand When a Sum is Even

Recall the rule for even sums:
- **Even + Even = Even**
- **Odd + Odd = Even**
- **Even + Odd = Odd**

So, for the sum of two numbers to be even, **both numbers must be even or both must be odd.**

---

### Step 3: Count the Total Number of Possible Outcomes

Each of Paul and Jesse chooses a number independently from the six primes. Thus, the total number of possible pairs is:
$$
6 \times 6 = 36
$$

---

### Step 4: Count the Number of Favorable Outcomes

#### Case 1: Both numbers are **even**

Only one even prime exists (i.e., 2), so both choosing the same even number is one possible pair:
$$
\text{Even + Even = 1 pair}
$$

#### Case 2: Both numbers are **odd**

There are 5 odd primes. So the number of favorable pairs where both choose odd:
$$
5 \times 5 = 25
$$

Total favorable outcomes:
$$
1 \text{ (even + even)} + 25 \text{ (odd + odd)} = 26
$$

---

### Step 5: Compute the Probability

$$
\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{26}{36}
$$

Simplifying:
$$
\frac{26}{36} = \frac{13}{18}
$$

---

### Final Answer

$$
\boxed{\dfrac{13}{18}}
$$

[Assistant]
Explanation:
Okay, so Paul and Jesse each choose a number at random from the first six primes. I need to find the probability that the sum of the numbers they choose is even. Hmm, let me think about how to approach this.

First, let me recall the first six prime numbers. The primes start at 2, right? So the first six primes are 2, 3, 5, 7, 11, and 13. Let me write those down:

First six primes: 2, 3, 5, 7, 11, 13.

Okay, so both Paul and Jesse are choosing one of these numbers at random. Assuming that each number is equally likely to be chosen, and their choices are independent, right? So each has 6 possible choices, so there are 6 x 6 = 36 total possible pairs of numbers they could choose.

Now, the question is about the probability that their sum is even. To find this probability, I need to figure out how many of these 36 pairs result in an even sum, and then divide that by 36.

So, probability = (number of favorable outcomes) / (total number of outcomes) = (number of pairs with even sum) / 36.

Alright, so how can the sum of two numbers be even? Well, I remember that even + even = even, odd + odd = even, and even + odd = odd. So, for the sum to be even, both numbers have to be even or both have to be odd.

Therefore, the problem reduces to finding the number of pairs where both numbers are even or both are odd. Then, adding those two counts together, and dividing by 36.

So, let me figure out how many even numbers and how many odd numbers are in the first six primes.

Looking at the list: 2, 3, 5, 7, 11, 13.

Let me check each number's parity:

- 2 is even.
- 3 is odd.
- 5 is odd.
- 7 is odd.
- 11 is odd.
- 13 is odd.

So, there's only one even prime number, which is 2. All the others are odd. Therefore, in the first six primes, there is 1 even number and 5 odd numbers.

Therefore, the number of ways both Paul and Jesse can pick even numbers: since there's only 1 even number, Paul can choose 2, and Jesse can also choose 2. So, how many such pairs? Well, since Paul has 1 choice and Jesse has 1 choice, so 1 x 1 = 1 pair where both numbers are even.

Wait, but hold on. Wait, if they are choosing numbers at random, does that mean that they can choose the same number? For example, can both Paul and Jesse choose 2? I think so, since there's no restriction mentioned. The problem says they choose a number at random from the first six primes. So repetition is allowed? Or is it without replacement? Wait, the problem doesn't specify that they can't choose the same number. So, I think it's with replacement. Therefore, the total number of possible pairs is indeed 6 x 6 = 36, and each number is equally likely for both Paul and Jesse.

Therefore, the number of pairs where both numbers are even is 1 (for Paul) times 1 (for Jesse) = 1 pair.

Now, the number of pairs where both numbers are odd. There are 5 odd numbers. So, Paul can choose any of the 5 odd numbers, and Jesse can also choose any of the 5 odd numbers. So, that's 5 x 5 = 25 pairs.

Therefore, total number of favorable outcomes is 1 + 25 = 26.

Therefore, the probability is 26/36, which simplifies to 13/18.

Wait, but let me check again. Let me think about whether that's correct.

Alternatively, maybe I can approach this problem by thinking about the probability that both numbers are even or both are odd.

So, probability that both are even: probability Paul picks even times probability Jesse picks even. Since there's 1 even out of 6, so (1/6)*(1/6) = 1/36.

Probability that both are odd: probability Paul picks odd times probability Jesse picks odd. There are 5 odd numbers, so (5/6)*(5/6) = 25/36.

Adding those two probabilities: 1/36 + 25/36 = 26/36 = 13/18. So, same result. Therefore, the probability is 13/18.

Hmm, that seems straightforward, but let me check if there's a possibility that I made an error in counting the number of even and odd primes.

Wait, primes are numbers greater than 1 that have no divisors other than 1 and themselves. The first prime is 2, which is even, and all other primes are odd because if they were even, they would be divisible by 2. So, indeed, except for 2, all primes are odd. Therefore, in the first six primes, only 2 is even, so 1 even, 5 odd. That seems correct.

Another way to think about it: The sum is even if both numbers are even or both are odd. Since there is only one even number, the number of ways both are even is 1*1=1. The number of ways both are odd is 5*5=25. So, total favorable is 26. Total possibilities 36. 26/36=13/18≈0.722...

Alternatively, maybe I can think of it as the probability that the sum is even is equal to the probability that both numbers are even plus the probability that both are odd. Which is exactly what I did.

Alternatively, maybe think in terms of parity. Let me think.

Let me note that the parity of the sum depends on the parities of the numbers. So, if I think of each person's number as either even or odd.

So, Paul's number can be even with probability 1/6, odd with probability 5/6. Same for Jesse.

Then, the probability that both are even is (1/6)(1/6) = 1/36, and both odd is (5/6)(5/6)=25/36. So, adding those gives the same 26/36.

Alternatively, maybe think of it as the sum is even if they have the same parity. Therefore, the probability that they have the same parity.

Alternatively, maybe compute the probability that they have different parities, and subtract from 1.

So, probability that sum is even = 1 - probability that sum is odd.

Probability that sum is odd is probability that one is even and the other is odd.

Which would be (probability Paul even and Jesse odd) + (probability Paul odd and Jesse even).

Which is (1/6 * 5/6) + (5/6 * 1/6) = 5/36 + 5/36 = 10/36.

Therefore, probability sum is even is 1 - 10/36 = 26/36 = 13/18.

Same result. So, that cross-check works.

Therefore, seems like 13/18 is the answer.

But let me think again, just to make sure. Maybe list out all possibilities? Although 36 possibilities is a bit tedious, but maybe check for some cases.

Wait, if there's only one even number, 2. So, the possible pairs where both are even: only (2,2). So, that's one pair.

Pairs where both are odd: there are 5 odd numbers, so 5x5=25 pairs. So, 25 pairs where both are odd.

Therefore, total of 26 pairs.

Total possible pairs: 36. So, 26/36 reduces to 13/18.

Alternatively, maybe there's a mistake here. Let me think: is there any possibility that when they pick numbers, they can't pick the same number? Wait, the problem says "choose a number at random from the first six primes." It doesn't specify whether they are choosing with replacement or without replacement.

Wait, but usually, when problems say "choose a number at random" without specifying, it's usually with replacement. But sometimes, it can be interpreted as choosing two different numbers. However, in this case, since it's about the sum, and the problem doesn't mention anything about distinct numbers, it's safer to assume that they can choose the same number. For example, if they were picking two different numbers, the total number of possibilities would be 6*5=30. However, since the problem didn't specify distinctness, and given that in most probability problems unless stated otherwise, selections are with replacement.

But wait, actually, if they are choosing numbers independently, then it's with replacement. So, for example, if Paul chooses a number and then Jesse chooses a number, they can choose the same number, because they are independent choices. Therefore, the total number of possibilities is 6*6=36.

Therefore, the initial reasoning holds. So, 26/36 is correct.

But just to be thorough, let me check if maybe the problem is from a source where "choosing a number at random" implies distinct numbers. But since it's not specified, I think with replacement is the standard assumption.

Alternatively, if we assume that they are choosing two distinct numbers, then the total number of possible outcomes would be 6*5=30. Then, how many pairs would be both even or both odd?

But in that case, since there is only one even number, there's no way to have two even numbers if they have to be distinct. Therefore, the number of pairs where both are even would be zero. Then, the number of pairs where both are odd would be C(5,2) = 10, but since order matters (since Paul and Jesse are distinct), it's 5*4=20. Therefore, total favorable outcomes would be 20, and total is 30. Therefore, probability would be 20/30 = 2/3 ≈ 0.666..., which is different from 13/18≈0.722.

But since the problem says "each choose a number at random from the first six primes", and doesn't mention anything about distinct numbers, it's more likely that they can choose the same number. Therefore, the answer is 13/18.

But just to make sure, let me check if there's any other possible mistake. For example, maybe the first six primes are different? Wait, first six primes are definitely 2, 3, 5, 7, 11, 13. Let me check that.

Primes are numbers greater than 1 that have no positive divisors other than 1 and themselves.

First prime: 2

Second: 3

Third: 5

Fourth: 7

Fifth: 11

Sixth: 13. Yeah, correct.

So, only 2 is even, rest are odd. Therefore, as before.

Another possible mistake: thinking that the sum is even requires both numbers even, but actually odd + odd is also even. So, that's correct.

So, with that, the number of favorable cases is 1 + 25 = 26.

Another check: 26 divided by 36. Simplify by dividing numerator and denominator by 2: 13/18.

Yes.

Alternatively, if I think in terms of probability, as I did before, 1/36 + 25/36 = 26/36 = 13/18.

Therefore, the answer should be 13/18.

Therefore, I think that's the correct answer.

**Final Answer**
The probability that the sum of the numbers they choose is even is \boxed{\dfrac{13}{18}}.

To determine the probability that the sum of the numbers chosen by Paul and Jesse is even, we begin by identifying all the numbers involved and analyzing the relevant properties of even and odd sums.

---

### Step 1: Identify the Numbers

The first six prime numbers are:
$$
\{2, 3, 5, 7, 11, 13\}
$$
Among these:
- **Even number:** 2 (only one is even)
- **Odd numbers:** 3, 5, 7, 11, 13 (five in total)

---

### Step 2: Understand When a Sum is Even

Recall the rule for even sums:
- **Even + Even = Even**
- **Odd + Odd = Even**
- **Even + Odd = Odd**

So, for the sum of two numbers to be even, **both numbers must be even or both must be odd.**

---

### Step 3: Count the Total Number of Possible Outcomes

Each of Paul and Jesse chooses a number independently from the six primes. Thus, the total number of possible pairs is:
$$
6 \times 6 = 36
$$

---

### Step 4: Count the Number of Favorable Outcomes

#### Case 1: Both numbers are **even**

Only one even prime exists (i.e., 2), so both choosing the same even number is one possible pair:
$$
\text{Even + Even = 1 pair}
$$

#### Case 2: Both numbers are **odd**

There are 5 odd primes. So the number of favorable pairs where both choose odd:
$$
5 \times 5 = 25
$$

Total favorable outcomes:
$$
1 \text{ (even + even)} + 25 \text{ (odd + odd)} = 26
$$

---

### Step 5: Compute the Probability

$$
\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{26}{36}
$$

Simplifying:
$$
\frac{26}{36} = \frac{13}{18}
$$

---

### Final Answer

$$
\boxed{\dfrac{13}{18}}
$$

Solution: The only way for the sum of the numbers Paul and Jesse choose to be odd is if one of them chooses 2 and the other chooses an odd prime. There are five ways for Paul to choose 2 and Jesse to choose an odd prime, and there are five ways for Jesse to choose 2 and Paul to choose an odd prime. Since there are $6\cdot 6=36$ total possible ways for Paul and Jesse to choose their numbers, the probability that the sum of the numbers Paul and Jesse choose is NOT even is $\frac{10}{36}=\frac{5}{18}$. Therefore, the probability that the sum of the numbers Paul and Jesse choose IS even is $1-\frac{5}{18}=\boxed{\frac{13}{18}}$.


Explanation:
Okay, so I need to find the maximum value of the expression (x - y)/(x⁴ + y⁴ + 6) where x and y are real numbers. Hmm, let me think about how to approach this.

First, I know that maximizing a function of two variables can sometimes be done using calculus, by taking partial derivatives and setting them to zero. But maybe there's a smarter way? Maybe symmetry or substitution? Let me see.

The denominator is x⁴ + y⁴ + 6. The numerator is x - y. Since both x and y are real numbers, their fourth powers are always non-negative. So the denominator is always at least 6, right? Because x⁴ and y⁴ are both ≥ 0, so x⁴ + y⁴ ≥ 0, so adding 6 makes it ≥ 6. Therefore, the denominator is always positive, so the expression is defined for all real x and y, and we don't have to worry about division by zero.

So, the expression is (x - y)/something that's always ≥ 6. So maybe the maximum occurs when the numerator is as large as possible and the denominator is as small as possible? But how do these two relate?

Alternatively, perhaps I can use some inequality to bound the expression. Or maybe set variables such that I can reduce the problem to one variable?

Let me consider substituting variables. Maybe let u = x + y and v = x - y? Wait, but the numerator is x - y, so if I set v = x - y, then the numerator is v. But then how does that relate to the denominator?

Alternatively, maybe set t = x - y. Then, we need to express the denominator in terms of t? But I don't know if that helps directly. Let's think.

Alternatively, maybe set y = kx? If I can express y in terms of x, then maybe I can write the function in terms of x and k? Let's try that.

Let me assume that y = kx, where k is some real number. Then the expression becomes (x - kx)/(x⁴ + (kx)^4 + 6) = x(1 - k)/(x⁴(1 + k⁴) + 6). Hmm, maybe I can set this as a function of x, and then find its maximum?

But then x can be positive or negative. Let me consider substituting t = x², but not sure. Alternatively, maybe think about the expression as a function of x, but for fixed k.

Alternatively, maybe take derivative with respect to x? Let me try that.

Let me denote f(x) = [x(1 - k)] / [x⁴(1 + k⁴) + 6]. Then, take the derivative of f(x) with respect to x. Then set to zero. But this might get complicated. Maybe there's a better approach.

Alternatively, maybe consider symmetry. Suppose we set y = -x? Then the numerator is x - (-x) = 2x. The denominator becomes x⁴ + (-x)^4 + 6 = 2x⁴ + 6. Then the expression is 2x/(2x⁴ + 6) = x/(x⁴ + 3). Then, maybe find the maximum of that function. Hmm, but is this the maximum? Not sure, maybe. But maybe it's a way to get some bounds.

Alternatively, maybe set x = y? Then the numerator is zero, so the expression is zero. Not helpful for maximum.

Alternatively, maybe set y to be something related to x. Suppose I fix the difference x - y = t. Then, perhaps express y in terms of x as y = x - t, then substitute into the denominator. Let me try that.

Let me set y = x - t. Then the expression becomes:

Numerator: t

Denominator: x⁴ + (x - t)^4 + 6

So, the expression is t / [x⁴ + (x - t)^4 + 6]. Now, perhaps we can think of this as for fixed t, we can choose x to minimize the denominator? Because the expression is t divided by something, so to maximize the expression for fixed t, we need to minimize the denominator. But how?

Alternatively, maybe for each t, find the minimum of x⁴ + (x - t)^4 + 6, then take the maximum over t of t divided by that minimum. This might be complicated, but maybe for some t, this gives us the maximum.

Alternatively, maybe set variables such that we can use calculus. Let's get back to the original function: f(x, y) = (x - y)/(x⁴ + y⁴ + 6). To find its maximum, we can take partial derivatives with respect to x and y, set them equal to zero, and solve the equations.

Let me try that. Let me denote the function as f(x, y). Then the partial derivatives are:

df/dx = [ (1)(x⁴ + y⁴ + 6) - (x - y)(4x³) ] / (x⁴ + y⁴ + 6)^2

Similarly,

df/dy = [ (-1)(x⁴ + y⁴ + 6) - (x - y)(4y³) ] / (x⁴ + y⁴ + 6)^2

Set both partial derivatives equal to zero. Since the denominators are positive, we can set the numerators equal to zero:

For df/dx = 0:

(x⁴ + y⁴ + 6) - 4x³(x - y) = 0

Similarly, for df/dy = 0:

- (x⁴ + y⁴ + 6) - 4y³(x - y) = 0

So, we have two equations:

1) x⁴ + y⁴ + 6 - 4x³(x - y) = 0

2) -x⁴ - y⁴ - 6 - 4y³(x - y) = 0

Let me write them again:

Equation (1): x⁴ + y⁴ + 6 = 4x³(x - y)

Equation (2): -x⁴ - y⁴ - 6 = 4y³(x - y)

If we add equations (1) and (2), we get:

(x⁴ + y⁴ + 6) + (-x⁴ - y⁴ - 6) = 4x³(x - y) + 4y³(x - y)

Simplify left side: 0

Right side: 4(x - y)(x³ + y³)

Therefore:

0 = 4(x - y)(x³ + y³)

Therefore, either x - y = 0 or x³ + y³ = 0.

But if x - y = 0, then x = y, which makes the numerator zero. So f(x, y) = 0, which is not the maximum. So the other case must hold: x³ + y³ = 0.

So, x³ + y³ = 0 => y³ = -x³ => y = -x (since cube roots are unique in real numbers). Therefore, y = -x.

So, now, substitute y = -x into one of the equations. Let me take equation (1):

x⁴ + y⁴ + 6 = 4x³(x - y)

Substituting y = -x:

x⁴ + (-x)^4 + 6 = 4x³(x - (-x)) => x⁴ + x⁴ + 6 = 4x³(2x) => 2x⁴ + 6 = 8x⁴

Simplify:

2x⁴ + 6 = 8x⁴ => 6 = 6x⁴ => x⁴ = 1 => x = ±1

Therefore, x = 1 or x = -1, and y = -x, so y = -1 or y = 1, respectively.

Therefore, we have critical points at (1, -1) and (-1, 1). Let's compute the value of the function at these points.

At (1, -1):

(1 - (-1))/(1^4 + (-1)^4 + 6) = (2)/(1 + 1 + 6) = 2/8 = 1/4

At (-1, 1):

(-1 - 1)/[(-1)^4 + (1)^4 + 6] = (-2)/(1 + 1 + 6) = -2/8 = -1/4

So, the function is 1/4 at (1, -1) and -1/4 at (-1, 1). Since we are looking for the maximum value, 1/4 is a candidate. But we need to check if this is indeed the maximum, or if there are other critical points or if the maximum occurs at infinity?

Wait, but when we did the partial derivatives, we found critical points only when y = -x. But maybe there are other critical points? Wait, we considered that when adding equations (1) and (2), we found that either x - y = 0 or x³ + y³ = 0. But x - y = 0 leads to zero function value, so we can ignore that. So the only critical points are when y = -x, leading to x = ±1. So maybe 1/4 is the maximum? But maybe there are other points where the function could be larger?

Alternatively, we need to check the limits as x or y go to infinity. Let me see.

Suppose x approaches infinity and y is fixed. Then the numerator is x - y ~ x, denominator ~ x⁴. So the whole expression ~ x / x⁴ = 1/x³ → 0.

Similarly, if y approaches infinity and x is fixed, then numerator is x - y ~ -y, denominator ~ y⁴. So the expression ~ -y / y⁴ = -1/y³ → 0.

If both x and y go to infinity, but in some relation. For example, suppose x approaches infinity and y = kx, where k is some constant. Then:

Numerator is x - y = x - kx = x(1 - k)

Denominator is x^4 + (kx)^4 + 6 = x^4(1 + k^4) + 6 ~ x^4(1 + k^4) as x approaches infinity.

Thus, the expression ~ [x(1 - k)] / [x^4(1 + k^4)] = (1 - k)/[x³(1 + k^4)] → 0 as x → infinity.

So, in all these cases, the limit is zero. Therefore, the maximum is likely to be achieved at some finite point, and we found that 1/4 is achieved at (1, -1). But we need to check if there are other critical points or if maybe the maximum could be higher somewhere else.

Wait, but we found that the only critical points are (1, -1) and (-1, 1). But maybe there are other critical points? Let me check.

Wait, the equations came from setting partial derivatives to zero. We assumed that x - y ≠ 0, leading to y = -x. But maybe there's a case where both equations hold even if x - y = 0? Wait, if x - y = 0, then numerator of the function is zero, so unless the denominator is also zero, but denominator is always ≥ 6. So, the function is zero at those points. So, they are not maxima. So, the critical points are only the ones where y = -x.

Therefore, maybe 1/4 is the maximum. But let me verify with another approach.

Another approach: Let me consider substituting variables to reduce the problem. Let me set t = x - y. Then, we need to maximize t/(x⁴ + y⁴ + 6). But how to relate t with x and y?

Alternatively, maybe use Cauchy-Schwarz inequality or other inequalities? Let me think.

Alternatively, let me consider that x⁴ + y⁴ ≥ (x² + y²)² / 2 by Cauchy-Schwarz? Wait, actually, there's an inequality that x⁴ + y⁴ ≥ (x² + y²)² / 2. Wait, let me check. Let me recall that for any real numbers a and b, (a² + b²)² ≤ 2(a⁴ + b⁴). Wait, that's the inequality. So, x⁴ + y⁴ ≥ (x² + y²)² / 2. Therefore, the denominator is x⁴ + y⁴ + 6 ≥ (x² + y²)² / 2 + 6. So, perhaps that can help?

But how does that relate to the numerator? The numerator is x - y. Maybe relate x - y to x² + y²?

Alternatively, perhaps use the Cauchy-Schwarz inequality on the numerator. Let me think. If I consider vectors (1, 1) and (x - y, something). Not sure. Alternatively, maybe use Hölder's inequality?

Alternatively, since the denominator is symmetric in x and y (except for the numerator), but if we consider that maybe the maximum occurs when y = -x, which we already considered. Since we found that at y = -x, the expression is 1/4. Maybe that's the maximum?

Alternatively, let me suppose that x and y are such that x - y is positive, because we want to maximize the expression. So let's suppose that x > y, so that the numerator is positive. Let me consider variables substitution. Let me set u = x + y and v = x - y. Then, we can write x = (u + v)/2, y = (u - v)/2. Let me try this substitution.

So, u = x + y, v = x - y. Then, x = (u + v)/2, y = (u - v)/2. Then x⁴ + y⁴ can be written in terms of u and v.

First, compute x⁴ + y⁴. Let me compute x^4 + y^4:

[(u + v)/2]^4 + [(u - v)/2]^4.

Let me expand that. Let me recall that (a + b)^4 + (a - b)^4 = 2(a^4 + 6a²b² + b^4). Wait, no, let me compute:

Let me compute (a + b)^4 + (a - b)^4:

= [a^4 + 4a³b + 6a²b² + 4ab³ + b^4] + [a^4 - 4a³b + 6a²b² - 4ab³ + b^4]

= 2a^4 + 12a²b² + 2b^4

Therefore, in our case, a = u/2, b = v/2. Wait, no. Wait, x = (u + v)/2, y = (u - v)/2. So, x and y are expressed in terms of u and v. So, to compute x^4 + y^4:

Let me denote x = (u + v)/2, y = (u - v)/2.

Then, x^4 + y^4 = [(u + v)/2]^4 + [(u - v)/2]^4 = [ (u^4 + 4u³v + 6u²v² + 4uv³ + v^4 ) + (u^4 - 4u³v + 6u²v² - 4uv³ + v^4) ] / 16

Which simplifies to [ 2u^4 + 12u²v² + 2v^4 ] / 16 = [ u^4 + 6u²v² + v^4 ] / 8

Therefore, the denominator is x⁴ + y⁴ + 6 = [u^4 + 6u²v² + v^4]/8 + 6.

So, the original expression becomes v / [ (u^4 + 6u²v² + v^4)/8 + 6 ] = 8v / [u^4 + 6u²v² + v^4 + 48].

So, we have the expression in terms of u and v: 8v / [u⁴ + 6u²v² + v⁴ + 48]. Now, since we are looking for maximum, and since v is the numerator, which we want to be as large as possible. However, u and v can be any real numbers. Wait, but perhaps we can find the relation between u and v to maximize this expression?

Alternatively, maybe set u = 0? Let me try that. If u = 0, then the denominator becomes 0 + 0 + v^4 + 48 = v^4 + 48, and the numerator is 8v. Therefore, expression is 8v / (v^4 + 48). Let's find the maximum of this function for real v.

Let me denote f(v) = 8v / (v⁴ + 48). To find its maximum, take derivative:

f’(v) = [8(v⁴ + 48) - 8v(4v³)] / (v⁴ + 48)^2 = [8v⁴ + 384 - 32v⁴] / denominator = [ -24v⁴ + 384 ] / denominator

Set numerator equal to zero:

-24v⁴ + 384 = 0 => 24v⁴ = 384 => v⁴ = 16 => v = ±2.

Therefore, critical points at v = 2 and v = -2. Compute f(2) = 8*2/(16 + 48) = 16/64 = 1/4. Similarly, f(-2) = -16/(16 + 48) = -16/64 = -1/4. So, maximum is 1/4 when u=0 and v=2. Which corresponds to u=0, v=2.

But u=0 and v=2 means that x + y = 0 and x - y = 2. Solving these: x + y = 0, x - y = 2. Adding equations: 2x = 2 => x = 1, then y = -1. Which is exactly the critical point we found earlier! So, that gives us 1/4. So, if we set u=0, we get the same maximum. But is there a possibility that for u ≠ 0, we can get a higher value?

So, to check that, let's consider if there exists u ≠ 0 such that the expression 8v / [u⁴ + 6u²v² + v⁴ + 48] is larger than 1/4. Let me analyze.

Let me consider that expression. Let me denote D = u⁴ + 6u²v² + v⁴ + 48. Let me try to see if for some u and v, the ratio 8v / D is larger than 1/4. Let me write:

8v / D > 1/4 ⇒ 32v > D ⇒ 32v > u⁴ + 6u²v² + v⁴ + 48.

This inequality needs to hold for some real u and v. Let me rearrange:

u⁴ + 6u²v² + v⁴ + 48 - 32v < 0.

But since u⁴ and v⁴ are non-negative, and other terms are quadratic or constants, maybe it's difficult to see. Alternatively, let me consider that for fixed v, can we choose u to minimize the denominator? Because 8v / D would be larger if D is smaller. So, for fixed v, minimize D(u) = u⁴ + 6u²v² + v⁴ + 48.

Let me do that. For fixed v, D(u) = u⁴ + 6u²v² + (v⁴ + 48). The term u⁴ + 6u²v² is minimized when u = 0, since it's a function of u: f(u) = u⁴ + 6u²v². Taking derivative with respect to u: 4u³ + 12u v². Setting to zero: u(4u² + 12v²) = 0. Solutions u = 0 or 4u² + 12v² = 0. Since u is real, the second equation implies u² = -3v², which only holds if u = v = 0. But if v ≠ 0, then the minimum occurs at u = 0. Therefore, the minimal D(u) for fixed v is when u = 0, which is v⁴ + 48. Therefore, the maximum of 8v / D is achieved when u = 0, as we considered before. Therefore, for any v, the maximum value of 8v / D is achieved when u = 0, hence the maximum of the original expression is 1/4.

Therefore, this suggests that 1/4 is indeed the maximum value. Therefore, the answer is 1/4. However, let me check another substitution to confirm.

Alternatively, suppose we set u ≠ 0. For example, take v = 2, and some u. Then D(u) = u⁴ + 6u²*(4) + 16 + 48 = u⁴ + 24u² + 64. Then, 8v / D(u) = 16 / (u⁴ + 24u² + 64). Let me see, the minimum of u⁴ + 24u² + 64 occurs when derivative is 4u³ + 48u = 0 ⇒ u(4u² + 48) = 0 ⇒ u = 0. Therefore, minimal D is 0 + 0 + 64 = 64, so the expression is 16/64 = 1/4. So same as before. If u is not zero, D increases, so 8v / D decreases. Therefore, indeed, maximum at u=0.

Therefore, the maximum value is 1/4. But let me check another case. Let me suppose v ≠ 0, but with u ≠ 0, maybe some combination gives higher value? For example, take u = something and v = something else. Let me try specific values.

Let me take x = 2, y = 0. Then numerator is 2 - 0 = 2. Denominator is 16 + 0 + 6 = 22. The value is 2 / 22 ≈ 0.09, which is less than 1/4 (which is 0.25). How about x = 1, y = 0? Then numerator is 1, denominator is 1 + 0 + 6 = 7, so 1/7 ≈ 0.14, still less than 0.25.

How about x = sqrt( something)? Wait, maybe trying specific points isn't helpful. Let me think of another way.

Alternatively, since we found via calculus that the only critical points are (1, -1) and (-1, 1), and they give 1/4 and -1/4, and we checked the limits, which go to zero, so likely that 1/4 is the maximum.

But let me check for another critical point. Suppose I consider the case where x and y are not related by y = -x. Wait, but according to the earlier analysis, when we set the partial derivatives to zero, the only solutions are when y = -x, except for the cases where x = y. But x = y gives zero. So perhaps there are no other critical points?

Alternatively, let me suppose that there is a maximum at some point not on y = -x. For example, take a point where x and y are such that the derivative conditions are satisfied. But according to the partial derivatives, the only possibilities are y = -x or x = y. Since x = y gives zero, which is not maximum, so only y = -x.

Therefore, the maximum is indeed 1/4. Therefore, the answer is \boxed{\dfrac{1}{4}}.

Wait, but let me just think again. Suppose we take x = t and y = -t. Then, the expression is (t - (-t))/(t⁴ + (-t)^4 + 6) = 2t/(2t⁴ + 6) = t/(t⁴ + 3). Let me find the maximum of t/(t⁴ + 3). Take derivative with respect to t: [1*(t⁴ + 3) - t*(4t³)] / (t⁴ + 3)^2 = [t⁴ + 3 - 4t⁴] / denominator = [ -3t⁴ + 3 ] / denominator. Setting numerator to zero: -3t⁴ + 3 = 0 ⇒ t⁴ = 1 ⇒ t = ±1. Then, for t = 1: 1/(1 + 3) = 1/4. For t = -1: -1/(1 + 3) = -1/4. So same result. So this also confirms that maximum is 1/4.

Therefore, all approaches lead to 1/4 as the maximum. Therefore, I think the answer is 1/4.

**Final Answer**
\boxed{\dfrac{1}{4}}

To find the **maximum value** of the expression
$$
\frac{x - y}{x^4 + y^4 + 6}
$$
over all real numbers $ x $ and $ y $, we proceed with a combination of analysis and calculus.

---

### Step 1: Domain Consideration

The denominator is $ x^4 + y^4 + 6 $. Since both $ x^4 $ and $ y^4 $ are non-negative, the denominator is always **positive** and at least 6. Therefore, the expression is **well-defined** for all real $ x $, $ y $, and the function is continuous everywhere.

---

### Step 2: Use of Symmetry

Let’s assume a substitution to simplify the expression. Let $ y = -x $. This gives:

- Numerator becomes $ x - y = x + x = 2x $
- Denominator becomes $ x^4 + (-x)^4 + 6 = x^4 + x^4 + 6 = 2x^4 + 6 $

So the expression simplifies to:
$$
\frac{2x}{2x^4 + 6} = \frac{x}{x^4 + 3}
$$

Now, we seek the **maximum** of $ f(x) = \frac{x}{x^4 + 3} $, which is a single-variable function.

---

### Step 3: Optimization (Calculus)

Find the critical points by differentiating:
$$
f(x) = \frac{x}{x^4 + 3}, \quad f'(x) = \frac{(x^4 + 3) - x \cdot 4x^3}{(x^4 + 3)^2} = \frac{x^4 + 3 - 4x^4}{(x^4 + 3)^2} = \frac{-3x^4 + 3}{(x^4 + 3)^2}
$$

Setting $ f'(x) = 0 $:
$$
-3x^4 + 3 = 0 \Rightarrow x^4 = 1 \Rightarrow x = \pm 1
$$

Now compute the function values at these critical points:

- For $ x = 1 $: $ f(1) = \frac{1}{1 + 3} = \frac{1}{4} $
- For $ x = -1 $: $ f(-1) = \frac{-1}{1 + 3} = -\frac{1}{4} $

So the maximum among these is $ \frac{1}{4} $, and the minimum is $ -\frac{1}{4} $.

---

### Step 4: Verification by Alternative Approach

Let’s consider the full two-variable function:
$$
f(x, y) = \frac{x - y}{x^4 + y^4 + 6}
$$

We can use symmetry or calculus by taking partial derivatives to confirm the critical points.

From calculus, solving the partial derivative equations leads to a condition $ x^3 + y^3 = 0 $, which implies $ y = -x $. Replacing this back into the original function simplifies it to the same function analyzed above.

This confirms that the critical point occurs at $ (x, y) = (1, -1) $ or $ (-1, 1) $, where the function attains **maximum value** of $ \frac{1}{4} $ and **minimum** of $ -\frac{1}{4} $.

---

### Step 5: Behavior at Infinity

To confirm the function does **not exceed** this maximum, note:

- As $ |x| $ or $ |y| \to \infty $, the numerator remains linear in $ x $ or $ y $, while the denominator grows rapidly with $ x^4 $ and $ y^4 $, so the overall expression tends to 0.
- Therefore, the function approaches zero at infinity and never exceeds the value at the critical point $ (1, -1) $.

---

### Final Conclusion

The **maximum value** of the expression
$$
\frac{x - y}{x^4 + y^4 + 6}
$$
over all real $ x $ and $ y $ is:

$$
\boxed{\dfrac{1}{4}}
$$

[Assistant]
Explanation:
Okay, so I need to find the maximum value of the expression (x - y)/(x⁴ + y⁴ + 6) where x and y are real numbers. Hmm, let me think about how to approach this.

First, I know that maximizing a function of two variables can sometimes be done using calculus, by taking partial derivatives and setting them to zero. But maybe there's a smarter way? Maybe symmetry or substitution? Let me see.

The denominator is x⁴ + y⁴ + 6. The numerator is x - y. Since both x and y are real numbers, their fourth powers are always non-negative. So the denominator is always at least 6, right? Because x⁴ and y⁴ are both ≥ 0, so x⁴ + y⁴ ≥ 0, so adding 6 makes it ≥ 6. Therefore, the denominator is always positive, so the expression is defined for all real x and y, and we don't have to worry about division by zero.

So, the expression is (x - y)/something that's always ≥ 6. So maybe the maximum occurs when the numerator is as large as possible and the denominator is as small as possible? But how do these two relate?

Alternatively, perhaps I can use some inequality to bound the expression. Or maybe set variables such that I can reduce the problem to one variable?

Let me consider substituting variables. Maybe let u = x + y and v = x - y? Wait, but the numerator is x - y, so if I set v = x - y, then the numerator is v. But then how does that relate to the denominator?

Alternatively, maybe set t = x - y. Then, we need to express the denominator in terms of t? But I don't know if that helps directly. Let's think.

Alternatively, maybe set y = kx? If I can express y in terms of x, then maybe I can write the function in terms of x and k? Let's try that.

Let me assume that y = kx, where k is some real number. Then the expression becomes (x - kx)/(x⁴ + (kx)^4 + 6) = x(1 - k)/(x⁴(1 + k⁴) + 6). Hmm, maybe I can set this as a function of x, and then find its maximum?

But then x can be positive or negative. Let me consider substituting t = x², but not sure. Alternatively, maybe think about the expression as a function of x, but for fixed k.

Alternatively, maybe take derivative with respect to x? Let me try that.

Let me denote f(x) = [x(1 - k)] / [x⁴(1 + k⁴) + 6]. Then, take the derivative of f(x) with respect to x. Then set to zero. But this might get complicated. Maybe there's a better approach.

Alternatively, maybe consider symmetry. Suppose we set y = -x? Then the numerator is x - (-x) = 2x. The denominator becomes x⁴ + (-x)^4 + 6 = 2x⁴ + 6. Then the expression is 2x/(2x⁴ + 6) = x/(x⁴ + 3). Then, maybe find the maximum of that function. Hmm, but is this the maximum? Not sure, maybe. But maybe it's a way to get some bounds.

Alternatively, maybe set x = y? Then the numerator is zero, so the expression is zero. Not helpful for maximum.

Alternatively, maybe set y to be something related to x. Suppose I fix the difference x - y = t. Then, perhaps express y in terms of x as y = x - t, then substitute into the denominator. Let me try that.

Let me set y = x - t. Then the expression becomes:

Numerator: t

Denominator: x⁴ + (x - t)^4 + 6

So, the expression is t / [x⁴ + (x - t)^4 + 6]. Now, perhaps we can think of this as for fixed t, we can choose x to minimize the denominator? Because the expression is t divided by something, so to maximize the expression for fixed t, we need to minimize the denominator. But how?

Alternatively, maybe for each t, find the minimum of x⁴ + (x - t)^4 + 6, then take the maximum over t of t divided by that minimum. This might be complicated, but maybe for some t, this gives us the maximum.

Alternatively, maybe set variables such that we can use calculus. Let's get back to the original function: f(x, y) = (x - y)/(x⁴ + y⁴ + 6). To find its maximum, we can take partial derivatives with respect to x and y, set them equal to zero, and solve the equations.

Let me try that. Let me denote the function as f(x, y). Then the partial derivatives are:

df/dx = [ (1)(x⁴ + y⁴ + 6) - (x - y)(4x³) ] / (x⁴ + y⁴ + 6)^2

Similarly,

df/dy = [ (-1)(x⁴ + y⁴ + 6) - (x - y)(4y³) ] / (x⁴ + y⁴ + 6)^2

Set both partial derivatives equal to zero. Since the denominators are positive, we can set the numerators equal to zero:

For df/dx = 0:

(x⁴ + y⁴ + 6) - 4x³(x - y) = 0

Similarly, for df/dy = 0:

- (x⁴ + y⁴ + 6) - 4y³(x - y) = 0

So, we have two equations:

1) x⁴ + y⁴ + 6 - 4x³(x - y) = 0

2) -x⁴ - y⁴ - 6 - 4y³(x - y) = 0

Let me write them again:

Equation (1): x⁴ + y⁴ + 6 = 4x³(x - y)

Equation (2): -x⁴ - y⁴ - 6 = 4y³(x - y)

If we add equations (1) and (2), we get:

(x⁴ + y⁴ + 6) + (-x⁴ - y⁴ - 6) = 4x³(x - y) + 4y³(x - y)

Simplify left side: 0

Right side: 4(x - y)(x³ + y³)

Therefore:

0 = 4(x - y)(x³ + y³)

Therefore, either x - y = 0 or x³ + y³ = 0.

But if x - y = 0, then x = y, which makes the numerator zero. So f(x, y) = 0, which is not the maximum. So the other case must hold: x³ + y³ = 0.

So, x³ + y³ = 0 => y³ = -x³ => y = -x (since cube roots are unique in real numbers). Therefore, y = -x.

So, now, substitute y = -x into one of the equations. Let me take equation (1):

x⁴ + y⁴ + 6 = 4x³(x - y)

Substituting y = -x:

x⁴ + (-x)^4 + 6 = 4x³(x - (-x)) => x⁴ + x⁴ + 6 = 4x³(2x) => 2x⁴ + 6 = 8x⁴

Simplify:

2x⁴ + 6 = 8x⁴ => 6 = 6x⁴ => x⁴ = 1 => x = ±1

Therefore, x = 1 or x = -1, and y = -x, so y = -1 or y = 1, respectively.

Therefore, we have critical points at (1, -1) and (-1, 1). Let's compute the value of the function at these points.

At (1, -1):

(1 - (-1))/(1^4 + (-1)^4 + 6) = (2)/(1 + 1 + 6) = 2/8 = 1/4

At (-1, 1):

(-1 - 1)/[(-1)^4 + (1)^4 + 6] = (-2)/(1 + 1 + 6) = -2/8 = -1/4

So, the function is 1/4 at (1, -1) and -1/4 at (-1, 1). Since we are looking for the maximum value, 1/4 is a candidate. But we need to check if this is indeed the maximum, or if there are other critical points or if the maximum occurs at infinity?

Wait, but when we did the partial derivatives, we found critical points only when y = -x. But maybe there are other critical points? Wait, we considered that when adding equations (1) and (2), we found that either x - y = 0 or x³ + y³ = 0. But x - y = 0 leads to zero function value, so we can ignore that. So the only critical points are when y = -x, leading to x = ±1. So maybe 1/4 is the maximum? But maybe there are other points where the function could be larger?

Alternatively, we need to check the limits as x or y go to infinity. Let me see.

Suppose x approaches infinity and y is fixed. Then the numerator is x - y ~ x, denominator ~ x⁴. So the whole expression ~ x / x⁴ = 1/x³ → 0.

Similarly, if y approaches infinity and x is fixed, then numerator is x - y ~ -y, denominator ~ y⁴. So the expression ~ -y / y⁴ = -1/y³ → 0.

If both x and y go to infinity, but in some relation. For example, suppose x approaches infinity and y = kx, where k is some constant. Then:

Numerator is x - y = x - kx = x(1 - k)

Denominator is x^4 + (kx)^4 + 6 = x^4(1 + k^4) + 6 ~ x^4(1 + k^4) as x approaches infinity.

Thus, the expression ~ [x(1 - k)] / [x^4(1 + k^4)] = (1 - k)/[x³(1 + k^4)] → 0 as x → infinity.

So, in all these cases, the limit is zero. Therefore, the maximum is likely to be achieved at some finite point, and we found that 1/4 is achieved at (1, -1). But we need to check if there are other critical points or if maybe the maximum could be higher somewhere else.

Wait, but we found that the only critical points are (1, -1) and (-1, 1). But maybe there are other critical points? Let me check.

Wait, the equations came from setting partial derivatives to zero. We assumed that x - y ≠ 0, leading to y = -x. But maybe there's a case where both equations hold even if x - y = 0? Wait, if x - y = 0, then numerator of the function is zero, so unless the denominator is also zero, but denominator is always ≥ 6. So, the function is zero at those points. So, they are not maxima. So, the critical points are only the ones where y = -x.

Therefore, maybe 1/4 is the maximum. But let me verify with another approach.

Another approach: Let me consider substituting variables to reduce the problem. Let me set t = x - y. Then, we need to maximize t/(x⁴ + y⁴ + 6). But how to relate t with x and y?

Alternatively, maybe use Cauchy-Schwarz inequality or other inequalities? Let me think.

Alternatively, let me consider that x⁴ + y⁴ ≥ (x² + y²)² / 2 by Cauchy-Schwarz? Wait, actually, there's an inequality that x⁴ + y⁴ ≥ (x² + y²)² / 2. Wait, let me check. Let me recall that for any real numbers a and b, (a² + b²)² ≤ 2(a⁴ + b⁴). Wait, that's the inequality. So, x⁴ + y⁴ ≥ (x² + y²)² / 2. Therefore, the denominator is x⁴ + y⁴ + 6 ≥ (x² + y²)² / 2 + 6. So, perhaps that can help?

But how does that relate to the numerator? The numerator is x - y. Maybe relate x - y to x² + y²?

Alternatively, perhaps use the Cauchy-Schwarz inequality on the numerator. Let me think. If I consider vectors (1, 1) and (x - y, something). Not sure. Alternatively, maybe use Hölder's inequality?

Alternatively, since the denominator is symmetric in x and y (except for the numerator), but if we consider that maybe the maximum occurs when y = -x, which we already considered. Since we found that at y = -x, the expression is 1/4. Maybe that's the maximum?

Alternatively, let me suppose that x and y are such that x - y is positive, because we want to maximize the expression. So let's suppose that x > y, so that the numerator is positive. Let me consider variables substitution. Let me set u = x + y and v = x - y. Then, we can write x = (u + v)/2, y = (u - v)/2. Let me try this substitution.

So, u = x + y, v = x - y. Then, x = (u + v)/2, y = (u - v)/2. Then x⁴ + y⁴ can be written in terms of u and v.

First, compute x⁴ + y⁴. Let me compute x^4 + y^4:

[(u + v)/2]^4 + [(u - v)/2]^4.

Let me expand that. Let me recall that (a + b)^4 + (a - b)^4 = 2(a^4 + 6a²b² + b^4). Wait, no, let me compute:

Let me compute (a + b)^4 + (a - b)^4:

= [a^4 + 4a³b + 6a²b² + 4ab³ + b^4] + [a^4 - 4a³b + 6a²b² - 4ab³ + b^4]

= 2a^4 + 12a²b² + 2b^4

Therefore, in our case, a = u/2, b = v/2. Wait, no. Wait, x = (u + v)/2, y = (u - v)/2. So, x and y are expressed in terms of u and v. So, to compute x^4 + y^4:

Let me denote x = (u + v)/2, y = (u - v)/2.

Then, x^4 + y^4 = [(u + v)/2]^4 + [(u - v)/2]^4 = [ (u^4 + 4u³v + 6u²v² + 4uv³ + v^4 ) + (u^4 - 4u³v + 6u²v² - 4uv³ + v^4) ] / 16

Which simplifies to [ 2u^4 + 12u²v² + 2v^4 ] / 16 = [ u^4 + 6u²v² + v^4 ] / 8

Therefore, the denominator is x⁴ + y⁴ + 6 = [u^4 + 6u²v² + v^4]/8 + 6.

So, the original expression becomes v / [ (u^4 + 6u²v² + v^4)/8 + 6 ] = 8v / [u^4 + 6u²v² + v^4 + 48].

So, we have the expression in terms of u and v: 8v / [u⁴ + 6u²v² + v⁴ + 48]. Now, since we are looking for maximum, and since v is the numerator, which we want to be as large as possible. However, u and v can be any real numbers. Wait, but perhaps we can find the relation between u and v to maximize this expression?

Alternatively, maybe set u = 0? Let me try that. If u = 0, then the denominator becomes 0 + 0 + v^4 + 48 = v^4 + 48, and the numerator is 8v. Therefore, expression is 8v / (v^4 + 48). Let's find the maximum of this function for real v.

Let me denote f(v) = 8v / (v⁴ + 48). To find its maximum, take derivative:

f’(v) = [8(v⁴ + 48) - 8v(4v³)] / (v⁴ + 48)^2 = [8v⁴ + 384 - 32v⁴] / denominator = [ -24v⁴ + 384 ] / denominator

Set numerator equal to zero:

-24v⁴ + 384 = 0 => 24v⁴ = 384 => v⁴ = 16 => v = ±2.

Therefore, critical points at v = 2 and v = -2. Compute f(2) = 8*2/(16 + 48) = 16/64 = 1/4. Similarly, f(-2) = -16/(16 + 48) = -16/64 = -1/4. So, maximum is 1/4 when u=0 and v=2. Which corresponds to u=0, v=2.

But u=0 and v=2 means that x + y = 0 and x - y = 2. Solving these: x + y = 0, x - y = 2. Adding equations: 2x = 2 => x = 1, then y = -1. Which is exactly the critical point we found earlier! So, that gives us 1/4. So, if we set u=0, we get the same maximum. But is there a possibility that for u ≠ 0, we can get a higher value?

So, to check that, let's consider if there exists u ≠ 0 such that the expression 8v / [u⁴ + 6u²v² + v⁴ + 48] is larger than 1/4. Let me analyze.

Let me consider that expression. Let me denote D = u⁴ + 6u²v² + v⁴ + 48. Let me try to see if for some u and v, the ratio 8v / D is larger than 1/4. Let me write:

8v / D > 1/4 ⇒ 32v > D ⇒ 32v > u⁴ + 6u²v² + v⁴ + 48.

This inequality needs to hold for some real u and v. Let me rearrange:

u⁴ + 6u²v² + v⁴ + 48 - 32v < 0.

But since u⁴ and v⁴ are non-negative, and other terms are quadratic or constants, maybe it's difficult to see. Alternatively, let me consider that for fixed v, can we choose u to minimize the denominator? Because 8v / D would be larger if D is smaller. So, for fixed v, minimize D(u) = u⁴ + 6u²v² + v⁴ + 48.

Let me do that. For fixed v, D(u) = u⁴ + 6u²v² + (v⁴ + 48). The term u⁴ + 6u²v² is minimized when u = 0, since it's a function of u: f(u) = u⁴ + 6u²v². Taking derivative with respect to u: 4u³ + 12u v². Setting to zero: u(4u² + 12v²) = 0. Solutions u = 0 or 4u² + 12v² = 0. Since u is real, the second equation implies u² = -3v², which only holds if u = v = 0. But if v ≠ 0, then the minimum occurs at u = 0. Therefore, the minimal D(u) for fixed v is when u = 0, which is v⁴ + 48. Therefore, the maximum of 8v / D is achieved when u = 0, as we considered before. Therefore, for any v, the maximum value of 8v / D is achieved when u = 0, hence the maximum of the original expression is 1/4.

Therefore, this suggests that 1/4 is indeed the maximum value. Therefore, the answer is 1/4. However, let me check another substitution to confirm.

Alternatively, suppose we set u ≠ 0. For example, take v = 2, and some u. Then D(u) = u⁴ + 6u²*(4) + 16 + 48 = u⁴ + 24u² + 64. Then, 8v / D(u) = 16 / (u⁴ + 24u² + 64). Let me see, the minimum of u⁴ + 24u² + 64 occurs when derivative is 4u³ + 48u = 0 ⇒ u(4u² + 48) = 0 ⇒ u = 0. Therefore, minimal D is 0 + 0 + 64 = 64, so the expression is 16/64 = 1/4. So same as before. If u is not zero, D increases, so 8v / D decreases. Therefore, indeed, maximum at u=0.

Therefore, the maximum value is 1/4. But let me check another case. Let me suppose v ≠ 0, but with u ≠ 0, maybe some combination gives higher value? For example, take u = something and v = something else. Let me try specific values.

Let me take x = 2, y = 0. Then numerator is 2 - 0 = 2. Denominator is 16 + 0 + 6 = 22. The value is 2 / 22 ≈ 0.09, which is less than 1/4 (which is 0.25). How about x = 1, y = 0? Then numerator is 1, denominator is 1 + 0 + 6 = 7, so 1/7 ≈ 0.14, still less than 0.25.

How about x = sqrt( something)? Wait, maybe trying specific points isn't helpful. Let me think of another way.

Alternatively, since we found via calculus that the only critical points are (1, -1) and (-1, 1), and they give 1/4 and -1/4, and we checked the limits, which go to zero, so likely that 1/4 is the maximum.

But let me check for another critical point. Suppose I consider the case where x and y are not related by y = -x. Wait, but according to the earlier analysis, when we set the partial derivatives to zero, the only solutions are when y = -x, except for the cases where x = y. But x = y gives zero. So perhaps there are no other critical points?

Alternatively, let me suppose that there is a maximum at some point not on y = -x. For example, take a point where x and y are such that the derivative conditions are satisfied. But according to the partial derivatives, the only possibilities are y = -x or x = y. Since x = y gives zero, which is not maximum, so only y = -x.

Therefore, the maximum is indeed 1/4. Therefore, the answer is \boxed{\dfrac{1}{4}}.

Wait, but let me just think again. Suppose we take x = t and y = -t. Then, the expression is (t - (-t))/(t⁴ + (-t)^4 + 6) = 2t/(2t⁴ + 6) = t/(t⁴ + 3). Let me find the maximum of t/(t⁴ + 3). Take derivative with respect to t: [1*(t⁴ + 3) - t*(4t³)] / (t⁴ + 3)^2 = [t⁴ + 3 - 4t⁴] / denominator = [ -3t⁴ + 3 ] / denominator. Setting numerator to zero: -3t⁴ + 3 = 0 ⇒ t⁴ = 1 ⇒ t = ±1. Then, for t = 1: 1/(1 + 3) = 1/4. For t = -1: -1/(1 + 3) = -1/4. So same result. So this also confirms that maximum is 1/4.

Therefore, all approaches lead to 1/4 as the maximum. Therefore, I think the answer is 1/4.

**Final Answer**
\boxed{\dfrac{1}{4}}

To find the **maximum value** of the expression
$$
\frac{x - y}{x^4 + y^4 + 6}
$$
over all real numbers $ x $ and $ y $, we proceed with a combination of analysis and calculus.

---

### Step 1: Domain Consideration

The denominator is $ x^4 + y^4 + 6 $. Since both $ x^4 $ and $ y^4 $ are non-negative, the denominator is always **positive** and at least 6. Therefore, the expression is **well-defined** for all real $ x $, $ y $, and the function is continuous everywhere.

---

### Step 2: Use of Symmetry

Let’s assume a substitution to simplify the expression. Let $ y = -x $. This gives:

- Numerator becomes $ x - y = x + x = 2x $
- Denominator becomes $ x^4 + (-x)^4 + 6 = x^4 + x^4 + 6 = 2x^4 + 6 $

So the expression simplifies to:
$$
\frac{2x}{2x^4 + 6} = \frac{x}{x^4 + 3}
$$

Now, we seek the **maximum** of $ f(x) = \frac{x}{x^4 + 3} $, which is a single-variable function.

---

### Step 3: Optimization (Calculus)

Find the critical points by differentiating:
$$
f(x) = \frac{x}{x^4 + 3}, \quad f'(x) = \frac{(x^4 + 3) - x \cdot 4x^3}{(x^4 + 3)^2} = \frac{x^4 + 3 - 4x^4}{(x^4 + 3)^2} = \frac{-3x^4 + 3}{(x^4 + 3)^2}
$$

Setting $ f'(x) = 0 $:
$$
-3x^4 + 3 = 0 \Rightarrow x^4 = 1 \Rightarrow x = \pm 1
$$

Now compute the function values at these critical points:

- For $ x = 1 $: $ f(1) = \frac{1}{1 + 3} = \frac{1}{4} $
- For $ x = -1 $: $ f(-1) = \frac{-1}{1 + 3} = -\frac{1}{4} $

So the maximum among these is $ \frac{1}{4} $, and the minimum is $ -\frac{1}{4} $.

---

### Step 4: Verification by Alternative Approach

Let’s consider the full two-variable function:
$$
f(x, y) = \frac{x - y}{x^4 + y^4 + 6}
$$

We can use symmetry or calculus by taking partial derivatives to confirm the critical points.

From calculus, solving the partial derivative equations leads to a condition $ x^3 + y^3 = 0 $, which implies $ y = -x $. Replacing this back into the original function simplifies it to the same function analyzed above.

This confirms that the critical point occurs at $ (x, y) = (1, -1) $ or $ (-1, 1) $, where the function attains **maximum value** of $ \frac{1}{4} $ and **minimum** of $ -\frac{1}{4} $.

---

### Step 5: Behavior at Infinity

To confirm the function does **not exceed** this maximum, note:

- As $ |x| $ or $ |y| \to \infty $, the numerator remains linear in $ x $ or $ y $, while the denominator grows rapidly with $ x^4 $ and $ y^4 $, so the overall expression tends to 0.
- Therefore, the function approaches zero at infinity and never exceeds the value at the critical point $ (1, -1) $.

---

### Final Conclusion

The **maximum value** of the expression
$$
\frac{x - y}{x^4 + y^4 + 6}
$$
over all real $ x $ and $ y $ is:

$$
\boxed{\dfrac{1}{4}}
$$

Solution:
Clearly, the maximum occurs when $x$ is positive and $y$ is negative.  Let $z = -y,$ so $z$ is positive, and $y = -z.$  Then
\[\frac{x - y}{x^4 + y^4 + 6} = \frac{x + z}{x^4 + z^4 + 6}.\]By AM-GM,
\[x^4 + 1 + 1 + 1 \ge 4 \sqrt[4]{x^4} = 4x,\]and
\[z^4 + 1 + 1 + 1 \ge 4 \sqrt[4]{z^4} = 4z.\]Then $x^4 + z^4 + 6 \ge 4(x + z),$ which implies
\[\frac{x + z}{x^4 + z^4 + 6} \le \frac{1}{4}.\]Equality occurs when $x = z = 1,$ so the maximum value is $\boxed{\frac{1}{4}}.$


Explanation:
Okay, so I need to figure out how many possible integer values of x there are such that x³ is a positive factor of 10!. Let me start by recalling what 10! is.

First, 10! means 10 factorial, which is 10 × 9 × 8 × ... × 1. Let me compute that value just to have it in mind. But maybe I don't need the exact number. Instead, I think I need the prime factorization of 10! because factors are related to prime factors.

Right, if I can find the prime factors of 10!, then I can figure out what exponents are possible for x³ to divide into it.

So, let me break down 10! into its prime factors. The primes less than or equal to 10 are 2, 3, 5, 7. Let me find the exponent of each prime in the factorization of 10!.

Starting with 2:

Number of 2s in 10! is calculated by floor(10/2) + floor(10/4) + floor(10/8) + ...

So that's 5 + 2 + 1 = 8.

For 3: floor(10/3) + floor(10/9) = 3 + 1 = 4.

For 5: floor(10/5) = 2.

For 7: floor(10/7) = 1.

So, 10! = 2⁸ × 3⁴ × 5² × 7¹.

Alright, so that's the prime factorization. Now, the problem says that x³ is a positive factor of 10!. So, if x³ divides 10!, then x must be such that when you cube it, all the exponents in its prime factorization are less than or equal to those in 10!.

So, if I denote x's prime factorization as 2^a × 3^b × 5^c × 7^d, then x³ would be 2^(3a) × 3^(3b) × 5^(3c) × 7^(3d). For this to divide 10! = 2⁸ × 3⁴ × 5² × 7¹, each exponent in x³ must be less than or equal to the corresponding exponent in 10!.

Therefore:

3a ≤ 8,

3b ≤ 4,

3c ≤ 2,

3d ≤ 1.

So, solving these inequalities for a, b, c, d.

Let me solve each one:

For a: 3a ≤ 8 => a ≤ 8/3 ≈ 2.666... Since a has to be an integer (because exponents in prime factors are integers), so the maximum a is 2.

Similarly, for b: 3b ≤ 4 => b ≤ 4/3 ≈ 1.333... So maximum b is 1.

For c: 3c ≤ 2 => c ≤ 2/3 ≈ 0.666... So maximum c is 0.

For d: 3d ≤ 1 => d ≤ 1/3 ≈ 0.333... So maximum d is 0.

Therefore, the exponents a, b, c, d can take the following values:

a: 0, 1, 2 (since 3*2=6 ≤8 and 3*3=9>8)

b: 0, 1 (since 3*1=3 ≤4 and 3*2=6>4)

c: 0 (since 3*0=0 ≤2 and 3*1=3>2)

d: 0 (similar reasoning, since 3*0=0 ≤1 and 3*1=3>1)

Therefore, the number of possible x is the product of the number of choices for each exponent.

So, for a: 3 choices (0,1,2)

For b: 2 choices (0,1)

For c: 1 choice (0)

For d: 1 choice (0)

Therefore, total number of x is 3 × 2 × 1 × 1 = 6.

Wait, but hold on. Let me check if that's all. Is there anything missing here?

Wait, x is an integer, so x can be 1, or other numbers. But does this count all possible x?

Wait, but according to this, the possible exponents are:

For 2: 0,1,2

For 3: 0,1

For 5: 0

For 7: 0

So, x is 2^a * 3^b * 5^c * 7^d, where a can be 0-2, b 0-1, c and d 0.

Therefore, the possible x's are:

When a=0, b=0: 1

a=0, b=1: 3

a=1, b=0: 2

a=1, b=1: 2*3=6

a=2, b=0: 4

a=2, b=1: 4*3=12

So that's 6 values: 1, 3, 2, 6, 4, 12.

Wait, that's 6 values, which matches with 3*2*1*1=6. But wait, is that all? Let me check if there are any other primes?

Wait, 10! only has primes up to 7. So x can't have any other prime factors, otherwise x³ would have primes not in 10! which would not divide 10!.

Therefore, x must be composed of primes 2, 3, 5, 7. But since exponents for 5 and 7 are zero, so x can't have 5 or 7 as factors. So, x must be a divisor of 2^2 * 3^1, since exponents for 2 can go up to 2 (so 2^2), and 3 can go up to 1. Therefore, x can be any divisor of 2² * 3¹.

Divisors of 2² * 3¹ are:

1, 2, 4, 3, 6, 12. Which are the ones I listed above. So 6 divisors. Therefore, the answer is 6?

Wait, but the problem says "possible integer values of x" where x³ divides 10!.

But wait, let me check if each of those x's indeed satisfy that x³ divides 10!.

Take x=1: 1³=1 divides anything. Yes.

x=2: 2³=8. 10! has 2⁸, so 8 divides 2⁸. Yes.

x=3: 3³=27. 10! has 3⁴. 27 divides 81? 81 divided by 27 is 3. So yes.

x=4: 4³=64. 2⁶. 10! has 2⁸, so yes.

x=6: 6³=216=2³×3³. Check if 2³×3³ divides 2⁸×3⁴. Yes, since 3 ≤8 and 3 ≤4.

x=12: 12³= (2²×3)^3= 2⁶×3³. Which divides 2⁸×3⁴. Yes, because 6 ≤8 and 3 ≤4.

So all of them are okay. Therefore, the answer is 6?

Wait, but hold on. Let me check if there are more possibilities. Because maybe there's a case where exponents can be higher?

Wait, let me re-examine the exponents. For example, for 5: if c is 0, but maybe 5 is allowed? Wait, no. Because if x has a factor of 5, then x³ would have 5³, but 10! only has 5². Therefore, 5³ cannot divide 5², so c has to be 0. Similarly for 7, d has to be 0. Therefore, x cannot have 5 or 7 as prime factors. So, x must be a divisor of 2^2 * 3^1. Which only has those 6 divisors as above.

But wait, hold on. Let me think again. Suppose x is 1, which is allowed. Also, 2, 3, 4, 6, 12. So that's 6 numbers. So the answer is 6?

But let me check if there's a mistake in my logic. For example, when I considered the exponents for 3: 3b ≤4. So b can be 0 or 1, since 3*1=3 ≤4, 3*2=6>4. So yes, only 0 or 1. Similarly, for 2: 3a ≤8. So a can be 0,1,2 because 3*2=6 ≤8, 3*3=9>8. So that's correct.

But wait, what if x is negative? The question says "positive factor" so x³ must be a positive factor. Since x³ is positive, x must be positive. Because if x is negative, x³ is negative, which can't divide 10! since factors are positive. Therefore, x must be a positive integer. So we don't have to consider negative integers, so our count of 6 is correct.

But let me check once again. Let me list all possible x:

Possible exponents for 2: 0,1,2

For 3: 0,1

So combinations:

(0,0): 2^0 *3^0=1

(0,1): 2^0*3^1=3

(1,0):2^1*3^0=2

(1,1):2*3=6

(2,0):4*1=4

(2,1):4*3=12

That's 6 numbers. So that's all. Therefore, the answer is 6.

Wait, but let me check if there are more possibilities. Let me think if there is a case where even if exponents are higher? For example, if I have a higher exponent for 2? Suppose a is 2. Then 3a is 6, which is ≤8. If a is 3, then 3*3=9 which is more than 8. So not allowed. So no. So a is up to 2. Similarly for others.

Alternatively, maybe I can think of the problem as follows: x³ | 10! => x | sqrt[3]{10!}. But since x must be integer, we need x to be such that x³ divides 10!.

But perhaps another way to approach this is to compute the exponents of primes in 10! and then for each prime, find the maximum exponent in x such that 3*exponent <= exponent in 10!.

Which is exactly what I did.

So, for each prime:

For prime 2: exponent in 10! is 8. So exponent in x can be at most floor(8/3) = 2.

For prime 3: exponent in 10! is 4. So floor(4/3)=1.

For prime 5: exponent in 10! is 2. floor(2/3)=0.

For prime 7: exponent in 10! is 1. floor(1/3)=0.

Therefore, the number of possible exponents is:

For prime 2: 2 - 0 + 1 = 3 (since exponents can be 0,1,2)

For prime 3: 1 - 0 +1 = 2 (0,1)

For prime 5: 0 - 0 +1 =1 (only 0)

For prime 7: 0 - 0 +1=1 (only 0)

Therefore, total number is 3*2*1*1=6. Same answer.

So seems like 6 is correct.

But wait, hold on. Wait, let me check if there is a possibility that x could have exponents that are non-integer? No, because exponents in prime factorization are integers. So x must have integer exponents, so the maximum exponents for each prime in x is floor( exponent_in_10! / 3 ).

So, seems correct.

Wait, but maybe there is a mistake here. Let me think again. Suppose someone might think that exponents can be higher if they are not multiples? But no. Because if x has exponent a for prime 2, then x³ has exponent 3a. So 3a must be less than or equal to 8. Therefore, a can be 0,1,2 because 3*2=6 <=8. If a=3, then 9>8. So that's not allowed.

Similarly for others. So yes, seems correct.

Alternatively, perhaps I need to think about the exponents of x. Let me consider that x is a positive integer. Then, x³ divides 10! if and only if, for each prime p, the exponent of p in x is at most floor(exponent of p in 10! / 3).

Therefore, for each prime, we compute floor(exponent / 3).

So for prime 2, exponent is 8, floor(8/3) = 2. So exponent of 2 in x can be 0,1,2.

For prime 3, exponent in 10! is 4, floor(4/3)=1, so exponent of 3 in x can be 0,1.

For prime 5, exponent in 10! is 2, floor(2/3)=0, so exponent of 5 in x can be 0.

Same for 7: floor(1/3)=0, exponent 0.

Therefore, the number of possible x is (2+1)*(1+1)*(0+1)*(0+1)= 3*2*1*1=6.

Therefore, same result. So seems like 6 is correct.

But to make sure, let me list all possible x:

As above: 1, 2, 3, 4, 6, 12.

Wait, but let me check if x= something else. For example, x= sqrt( something ). Wait, but x has to be integer. So, for example, if x= sqrt( something ) but that would not be integer.

Alternatively, is there a case where exponents can be something else?

Wait, for example, suppose x is 1, which is allowed. What about x= 2^2 * 3^1 = 4*3=12. Which is allowed.

Is there another number? Let me think. Let's check if x= 2^2 * 3^0=4, which is allowed.

What about x= 2^1*3^1=6, which is allowed.

Is there a case where exponents can be zero? For example, x=1, which is allowed.

Is there any other number? For instance, can x be 2^0 *3^0=1, yes.

What about x= 2^0 *3^1=3, yes.

Is there any other numbers? Let me think. Let me check if there is a number with higher exponents?

For example, if x is 2^3=8. Then x³= 8³=512. Is 512 divides 10! ?

But 10! is 3628800. Let me check if 512 divides that.

Well, prime factorization of 10! is 2^8 * 3^4 * 5^2 * 7.

So, 512 is 2^9. However, 10! only has 2^8. Therefore, 2^9 does not divide 2^8. Therefore, 8³ does not divide 10! Hence x=8 is invalid.

Similarly, x= 3^2=9. Then x³= 729=3^6. 10! has 3^4. 3^6 doesn't divide 3^4. Therefore, invalid.

Similarly, x=5. Then x³=125=5³. 10! has 5², so no.

x=7: 7³ doesn't divide 10!.

x= 2*5=10. Then x³= 1000=2³*5³. 10! has 2^8 and 5². So 5³ is too much. Therefore x=10 is invalid.

Similarly, x= 2*7=14. x³= 2³*7³. 7³ is too big.

x=3*5=15. x³=3³*5³. Again, 5³ too big.

x=3*7=21. x³=3³*7³. 7³ too big.

x=5*7=35. x³=... same problem.

x=2*3*5=30. x³= 2³*3³*5³. Again, 5³ too big.

x=2*3*7=42. Similarly, 7³ too big.

x=2*5*7=70. Same.

x=3*5*7=105. Same.

x=2*3*5*7=210. Same.

So, all numbers that include 5 or 7 will have their cube requiring exponents higher than allowed. Therefore, only numbers that are products of 2 and 3.

So, the possible x are numbers made up of 2^a * 3^b where a is 0,1,2 and b is 0,1. Which gives exactly the 6 numbers we found earlier.

Therefore, I think the answer is 6.

Wait, but hold on. Let me check another way. Let me compute the total number of divisors of 10! which are perfect cubes? No, the question is not about x being a cube, but x³ being a factor. So, different question.

But perhaps another way to think is, if x³ divides 10!, then x divides 10!^{1/3} ?

But since 10! isn't a perfect cube, but we can consider the exponents.

But maybe that's complicating.

Alternatively, since we have already verified through multiple approaches that the answer is 6, and that the numbers are 1, 2, 3, 4, 6, 12, which are all valid.

Therefore, unless there's a mistake in my logic, which I can't see, the answer is 6.

But wait, I just thought, what about x= something like 2^2 * 3^1 = 12, which is allowed. What about x= 2^1 * 3^0 = 2, allowed.

But is there a number like 2^0 * 3^0 =1. Yes.

Wait, perhaps another way to calculate the number of x: since x must be of the form 2^a * 3^b where 0 ≤ a ≤ 2 and 0 ≤ b ≤ 1.

So the number of possible x is (number of a's) * (number of b's) = 3 * 2 = 6.

Which is same as before.

Therefore, I think the answer is 6.

But let me check once more with actual numbers. Let me list all possible x and check:

1. x=1: 1³=1 divides 10!.

2. x=2: 8 divides 10!.

3. x=3: 27 divides 10!.

4. x=4: 64 divides 10!.

5. x=6: 216 divides 10!.

6. x=12: 1728 divides 10!?

Wait, let me check if 1728 divides 10!.

Wait, 10! = 3628800. Let me divide 3628800 by 1728.

First, compute 1728 × 2000 = 3,456,000. Which is more than 3,628,800? Wait:

Wait, 1728 × 2000 = 3,456,000

Subtract that from 3,628,800: 3,628,800 - 3,456,000 = 172,800

Now 1728 × 100 = 172,800. So total is 2000 + 100 = 2100. So 1728 × 2100 = 3,628,800. Which is exactly 10!.

Therefore, 1728 divides 10! So x=12 is okay.

So all six values of x are valid.

Therefore, I think the answer is 6. Hence, the number of possible integer values of x is 6.

But let me check once again if there is a number that I missed. For example, what about x= sqrt( something ). But x must be integer, so that is not possible.

Alternatively, if there are numbers with exponents that aren't covered? For example, could there be a number with exponents for 2 and 3 that are not in the ranges we considered?

No. Because we considered all possible exponents such that when multiplied by 3, they don't exceed the exponents in 10!.

Therefore, I think the answer is 6.

**Final Answer**
\boxed{6}

To solve the problem, we are given that $ x^3 $ is a **positive factor** of $ 10! $. Our goal is to determine how many **integer** values of $ x $ satisfy this condition.

---

### Step 1: Understand the Prime Factorization of 10!

We start by computing the prime factorization of $ 10! $:

$$
10! = 10 \times 9 \times 8 \times \dots \times 1
$$

We find the exponents of each prime number in the factorization:

- **2**: $ \left\lfloor \frac{10}{2} \right\rfloor + \left\lfloor \frac{10}{4} \right\rfloor + \left\lfloor \frac{10}{8} \right\rfloor = 5 + 2 + 1 = 8 $
- **3**: $ \left\lfloor \frac{10}{3} \right\rfloor + \left\lfloor \frac{10}{9} \right\rfloor = 3 + 1 = 4 $
- **5**: $ \left\lfloor \frac{10}{5} \right\rfloor = 2 $
- **7**: $ \left\lfloor \frac{10}{7} \right\rfloor = 1 $

So the prime factorization is:

$$
10! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7^1
$$

---

### Step 2: Use the Structure of $ x^3 \mid 10! $

We need $ x^3 $ to be a **factor** of $ 10! $, meaning that:

$$
x^3 \mid 10! \Rightarrow \text{each exponent of prime in } x^3 \leq \text{corresponding exponent in } 10!
$$

Let the prime factorization of $ x $ be:

$$
x = 2^a \cdot 3^b \cdot 5^c \cdot 7^d
$$

Then:

$$
x^3 = 2^{3a} \cdot 3^{3b} \cdot 5^{3c} \cdot 7^{3d}
$$

Now, impose the divisibility condition:

- $ 3a \leq 8 $ ⇒ $ a \leq \left\lfloor \frac{8}{3} \right\rfloor = 2 $
- $ 3b \leq 4 $ ⇒ $ b \leq \left\lfloor \frac{4}{3} \right\rfloor = 1 $
- $ 3c \leq 2 $ ⇒ $ c \leq \left\lfloor \frac{2}{3} \right\rfloor = 0 $
- $ 3d \leq 1 $ ⇒ $ d \leq \left\lfloor \frac{1}{3} \right\rfloor = 0 $

So, the possible exponents are:

- $ a \in \{0, 1, 2\} $ → 3 choices
- $ b \in \{0, 1\} $ → 2 choices
- $ c = 0 $, $ d = 0 $ → 1 choice each

---

### Step 3: Calculate Total Number of Valid $ x $

Each valid exponent combination gives a unique positive integer value of $ x $. The total number of values is:

$$
3 \times 2 \times 1 \times 1 = 6
$$

---

### Step 4: List All Valid $ x $ Values

Let’s enumerate them:

- $ a = 0, b = 0 $ ⇒ $ x = 1 $
- $ a = 0, b = 1 $ ⇒ $ x = 3 $
- $ a = 1, b = 0 $ ⇒ $ x = 2 $
- $ a = 1, b = 1 $ ⇒ $ x = 6 $
- $ a = 2, b = 0 $ ⇒ $ x = 4 $
- $ a = 2, b = 1 $ ⇒ $ x = 12 $

These values satisfy the condition that $ x^3 \mid 10! $. Any additional values would either involve primes not in $ 10! $ (like 5 or 7) or exponents that violate the divisibility condition.

---

### Final Answer

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many possible integer values of x there are such that x³ is a positive factor of 10!. Let me start by recalling what 10! is.

First, 10! means 10 factorial, which is 10 × 9 × 8 × ... × 1. Let me compute that value just to have it in mind. But maybe I don't need the exact number. Instead, I think I need the prime factorization of 10! because factors are related to prime factors.

Right, if I can find the prime factors of 10!, then I can figure out what exponents are possible for x³ to divide into it.

So, let me break down 10! into its prime factors. The primes less than or equal to 10 are 2, 3, 5, 7. Let me find the exponent of each prime in the factorization of 10!.

Starting with 2:

Number of 2s in 10! is calculated by floor(10/2) + floor(10/4) + floor(10/8) + ...

So that's 5 + 2 + 1 = 8.

For 3: floor(10/3) + floor(10/9) = 3 + 1 = 4.

For 5: floor(10/5) = 2.

For 7: floor(10/7) = 1.

So, 10! = 2⁸ × 3⁴ × 5² × 7¹.

Alright, so that's the prime factorization. Now, the problem says that x³ is a positive factor of 10!. So, if x³ divides 10!, then x must be such that when you cube it, all the exponents in its prime factorization are less than or equal to those in 10!.

So, if I denote x's prime factorization as 2^a × 3^b × 5^c × 7^d, then x³ would be 2^(3a) × 3^(3b) × 5^(3c) × 7^(3d). For this to divide 10! = 2⁸ × 3⁴ × 5² × 7¹, each exponent in x³ must be less than or equal to the corresponding exponent in 10!.

Therefore:

3a ≤ 8,

3b ≤ 4,

3c ≤ 2,

3d ≤ 1.

So, solving these inequalities for a, b, c, d.

Let me solve each one:

For a: 3a ≤ 8 => a ≤ 8/3 ≈ 2.666... Since a has to be an integer (because exponents in prime factors are integers), so the maximum a is 2.

Similarly, for b: 3b ≤ 4 => b ≤ 4/3 ≈ 1.333... So maximum b is 1.

For c: 3c ≤ 2 => c ≤ 2/3 ≈ 0.666... So maximum c is 0.

For d: 3d ≤ 1 => d ≤ 1/3 ≈ 0.333... So maximum d is 0.

Therefore, the exponents a, b, c, d can take the following values:

a: 0, 1, 2 (since 3*2=6 ≤8 and 3*3=9>8)

b: 0, 1 (since 3*1=3 ≤4 and 3*2=6>4)

c: 0 (since 3*0=0 ≤2 and 3*1=3>2)

d: 0 (similar reasoning, since 3*0=0 ≤1 and 3*1=3>1)

Therefore, the number of possible x is the product of the number of choices for each exponent.

So, for a: 3 choices (0,1,2)

For b: 2 choices (0,1)

For c: 1 choice (0)

For d: 1 choice (0)

Therefore, total number of x is 3 × 2 × 1 × 1 = 6.

Wait, but hold on. Let me check if that's all. Is there anything missing here?

Wait, x is an integer, so x can be 1, or other numbers. But does this count all possible x?

Wait, but according to this, the possible exponents are:

For 2: 0,1,2

For 3: 0,1

For 5: 0

For 7: 0

So, x is 2^a * 3^b * 5^c * 7^d, where a can be 0-2, b 0-1, c and d 0.

Therefore, the possible x's are:

When a=0, b=0: 1

a=0, b=1: 3

a=1, b=0: 2

a=1, b=1: 2*3=6

a=2, b=0: 4

a=2, b=1: 4*3=12

So that's 6 values: 1, 3, 2, 6, 4, 12.

Wait, that's 6 values, which matches with 3*2*1*1=6. But wait, is that all? Let me check if there are any other primes?

Wait, 10! only has primes up to 7. So x can't have any other prime factors, otherwise x³ would have primes not in 10! which would not divide 10!.

Therefore, x must be composed of primes 2, 3, 5, 7. But since exponents for 5 and 7 are zero, so x can't have 5 or 7 as factors. So, x must be a divisor of 2^2 * 3^1, since exponents for 2 can go up to 2 (so 2^2), and 3 can go up to 1. Therefore, x can be any divisor of 2² * 3¹.

Divisors of 2² * 3¹ are:

1, 2, 4, 3, 6, 12. Which are the ones I listed above. So 6 divisors. Therefore, the answer is 6?

Wait, but the problem says "possible integer values of x" where x³ divides 10!.

But wait, let me check if each of those x's indeed satisfy that x³ divides 10!.

Take x=1: 1³=1 divides anything. Yes.

x=2: 2³=8. 10! has 2⁸, so 8 divides 2⁸. Yes.

x=3: 3³=27. 10! has 3⁴. 27 divides 81? 81 divided by 27 is 3. So yes.

x=4: 4³=64. 2⁶. 10! has 2⁸, so yes.

x=6: 6³=216=2³×3³. Check if 2³×3³ divides 2⁸×3⁴. Yes, since 3 ≤8 and 3 ≤4.

x=12: 12³= (2²×3)^3= 2⁶×3³. Which divides 2⁸×3⁴. Yes, because 6 ≤8 and 3 ≤4.

So all of them are okay. Therefore, the answer is 6?

Wait, but hold on. Let me check if there are more possibilities. Because maybe there's a case where exponents can be higher?

Wait, let me re-examine the exponents. For example, for 5: if c is 0, but maybe 5 is allowed? Wait, no. Because if x has a factor of 5, then x³ would have 5³, but 10! only has 5². Therefore, 5³ cannot divide 5², so c has to be 0. Similarly for 7, d has to be 0. Therefore, x cannot have 5 or 7 as prime factors. So, x must be a divisor of 2^2 * 3^1. Which only has those 6 divisors as above.

But wait, hold on. Let me think again. Suppose x is 1, which is allowed. Also, 2, 3, 4, 6, 12. So that's 6 numbers. So the answer is 6?

But let me check if there's a mistake in my logic. For example, when I considered the exponents for 3: 3b ≤4. So b can be 0 or 1, since 3*1=3 ≤4, 3*2=6>4. So yes, only 0 or 1. Similarly, for 2: 3a ≤8. So a can be 0,1,2 because 3*2=6 ≤8, 3*3=9>8. So that's correct.

But wait, what if x is negative? The question says "positive factor" so x³ must be a positive factor. Since x³ is positive, x must be positive. Because if x is negative, x³ is negative, which can't divide 10! since factors are positive. Therefore, x must be a positive integer. So we don't have to consider negative integers, so our count of 6 is correct.

But let me check once again. Let me list all possible x:

Possible exponents for 2: 0,1,2

For 3: 0,1

So combinations:

(0,0): 2^0 *3^0=1

(0,1): 2^0*3^1=3

(1,0):2^1*3^0=2

(1,1):2*3=6

(2,0):4*1=4

(2,1):4*3=12

That's 6 numbers. So that's all. Therefore, the answer is 6.

Wait, but let me check if there are more possibilities. Let me think if there is a case where even if exponents are higher? For example, if I have a higher exponent for 2? Suppose a is 2. Then 3a is 6, which is ≤8. If a is 3, then 3*3=9 which is more than 8. So not allowed. So no. So a is up to 2. Similarly for others.

Alternatively, maybe I can think of the problem as follows: x³ | 10! => x | sqrt[3]{10!}. But since x must be integer, we need x to be such that x³ divides 10!.

But perhaps another way to approach this is to compute the exponents of primes in 10! and then for each prime, find the maximum exponent in x such that 3*exponent <= exponent in 10!.

Which is exactly what I did.

So, for each prime:

For prime 2: exponent in 10! is 8. So exponent in x can be at most floor(8/3) = 2.

For prime 3: exponent in 10! is 4. So floor(4/3)=1.

For prime 5: exponent in 10! is 2. floor(2/3)=0.

For prime 7: exponent in 10! is 1. floor(1/3)=0.

Therefore, the number of possible exponents is:

For prime 2: 2 - 0 + 1 = 3 (since exponents can be 0,1,2)

For prime 3: 1 - 0 +1 = 2 (0,1)

For prime 5: 0 - 0 +1 =1 (only 0)

For prime 7: 0 - 0 +1=1 (only 0)

Therefore, total number is 3*2*1*1=6. Same answer.

So seems like 6 is correct.

But wait, hold on. Wait, let me check if there is a possibility that x could have exponents that are non-integer? No, because exponents in prime factorization are integers. So x must have integer exponents, so the maximum exponents for each prime in x is floor( exponent_in_10! / 3 ).

So, seems correct.

Wait, but maybe there is a mistake here. Let me think again. Suppose someone might think that exponents can be higher if they are not multiples? But no. Because if x has exponent a for prime 2, then x³ has exponent 3a. So 3a must be less than or equal to 8. Therefore, a can be 0,1,2 because 3*2=6 <=8. If a=3, then 9>8. So that's not allowed.

Similarly for others. So yes, seems correct.

Alternatively, perhaps I need to think about the exponents of x. Let me consider that x is a positive integer. Then, x³ divides 10! if and only if, for each prime p, the exponent of p in x is at most floor(exponent of p in 10! / 3).

Therefore, for each prime, we compute floor(exponent / 3).

So for prime 2, exponent is 8, floor(8/3) = 2. So exponent of 2 in x can be 0,1,2.

For prime 3, exponent in 10! is 4, floor(4/3)=1, so exponent of 3 in x can be 0,1.

For prime 5, exponent in 10! is 2, floor(2/3)=0, so exponent of 5 in x can be 0.

Same for 7: floor(1/3)=0, exponent 0.

Therefore, the number of possible x is (2+1)*(1+1)*(0+1)*(0+1)= 3*2*1*1=6.

Therefore, same result. So seems like 6 is correct.

But to make sure, let me list all possible x:

As above: 1, 2, 3, 4, 6, 12.

Wait, but let me check if x= something else. For example, x= sqrt( something ). Wait, but x has to be integer. So, for example, if x= sqrt( something ) but that would not be integer.

Alternatively, is there a case where exponents can be something else?

Wait, for example, suppose x is 1, which is allowed. What about x= 2^2 * 3^1 = 4*3=12. Which is allowed.

Is there another number? Let me think. Let's check if x= 2^2 * 3^0=4, which is allowed.

What about x= 2^1*3^1=6, which is allowed.

Is there a case where exponents can be zero? For example, x=1, which is allowed.

Is there any other number? For instance, can x be 2^0 *3^0=1, yes.

What about x= 2^0 *3^1=3, yes.

Is there any other numbers? Let me think. Let me check if there is a number with higher exponents?

For example, if x is 2^3=8. Then x³= 8³=512. Is 512 divides 10! ?

But 10! is 3628800. Let me check if 512 divides that.

Well, prime factorization of 10! is 2^8 * 3^4 * 5^2 * 7.

So, 512 is 2^9. However, 10! only has 2^8. Therefore, 2^9 does not divide 2^8. Therefore, 8³ does not divide 10! Hence x=8 is invalid.

Similarly, x= 3^2=9. Then x³= 729=3^6. 10! has 3^4. 3^6 doesn't divide 3^4. Therefore, invalid.

Similarly, x=5. Then x³=125=5³. 10! has 5², so no.

x=7: 7³ doesn't divide 10!.

x= 2*5=10. Then x³= 1000=2³*5³. 10! has 2^8 and 5². So 5³ is too much. Therefore x=10 is invalid.

Similarly, x= 2*7=14. x³= 2³*7³. 7³ is too big.

x=3*5=15. x³=3³*5³. Again, 5³ too big.

x=3*7=21. x³=3³*7³. 7³ too big.

x=5*7=35. x³=... same problem.

x=2*3*5=30. x³= 2³*3³*5³. Again, 5³ too big.

x=2*3*7=42. Similarly, 7³ too big.

x=2*5*7=70. Same.

x=3*5*7=105. Same.

x=2*3*5*7=210. Same.

So, all numbers that include 5 or 7 will have their cube requiring exponents higher than allowed. Therefore, only numbers that are products of 2 and 3.

So, the possible x are numbers made up of 2^a * 3^b where a is 0,1,2 and b is 0,1. Which gives exactly the 6 numbers we found earlier.

Therefore, I think the answer is 6.

Wait, but hold on. Let me check another way. Let me compute the total number of divisors of 10! which are perfect cubes? No, the question is not about x being a cube, but x³ being a factor. So, different question.

But perhaps another way to think is, if x³ divides 10!, then x divides 10!^{1/3} ?

But since 10! isn't a perfect cube, but we can consider the exponents.

But maybe that's complicating.

Alternatively, since we have already verified through multiple approaches that the answer is 6, and that the numbers are 1, 2, 3, 4, 6, 12, which are all valid.

Therefore, unless there's a mistake in my logic, which I can't see, the answer is 6.

But wait, I just thought, what about x= something like 2^2 * 3^1 = 12, which is allowed. What about x= 2^1 * 3^0 = 2, allowed.

But is there a number like 2^0 * 3^0 =1. Yes.

Wait, perhaps another way to calculate the number of x: since x must be of the form 2^a * 3^b where 0 ≤ a ≤ 2 and 0 ≤ b ≤ 1.

So the number of possible x is (number of a's) * (number of b's) = 3 * 2 = 6.

Which is same as before.

Therefore, I think the answer is 6.

But let me check once more with actual numbers. Let me list all possible x and check:

1. x=1: 1³=1 divides 10!.

2. x=2: 8 divides 10!.

3. x=3: 27 divides 10!.

4. x=4: 64 divides 10!.

5. x=6: 216 divides 10!.

6. x=12: 1728 divides 10!?

Wait, let me check if 1728 divides 10!.

Wait, 10! = 3628800. Let me divide 3628800 by 1728.

First, compute 1728 × 2000 = 3,456,000. Which is more than 3,628,800? Wait:

Wait, 1728 × 2000 = 3,456,000

Subtract that from 3,628,800: 3,628,800 - 3,456,000 = 172,800

Now 1728 × 100 = 172,800. So total is 2000 + 100 = 2100. So 1728 × 2100 = 3,628,800. Which is exactly 10!.

Therefore, 1728 divides 10! So x=12 is okay.

So all six values of x are valid.

Therefore, I think the answer is 6. Hence, the number of possible integer values of x is 6.

But let me check once again if there is a number that I missed. For example, what about x= sqrt( something ). But x must be integer, so that is not possible.

Alternatively, if there are numbers with exponents that aren't covered? For example, could there be a number with exponents for 2 and 3 that are not in the ranges we considered?

No. Because we considered all possible exponents such that when multiplied by 3, they don't exceed the exponents in 10!.

Therefore, I think the answer is 6.

**Final Answer**
\boxed{6}

To solve the problem, we are given that $ x^3 $ is a **positive factor** of $ 10! $. Our goal is to determine how many **integer** values of $ x $ satisfy this condition.

---

### Step 1: Understand the Prime Factorization of 10!

We start by computing the prime factorization of $ 10! $:

$$
10! = 10 \times 9 \times 8 \times \dots \times 1
$$

We find the exponents of each prime number in the factorization:

- **2**: $ \left\lfloor \frac{10}{2} \right\rfloor + \left\lfloor \frac{10}{4} \right\rfloor + \left\lfloor \frac{10}{8} \right\rfloor = 5 + 2 + 1 = 8 $
- **3**: $ \left\lfloor \frac{10}{3} \right\rfloor + \left\lfloor \frac{10}{9} \right\rfloor = 3 + 1 = 4 $
- **5**: $ \left\lfloor \frac{10}{5} \right\rfloor = 2 $
- **7**: $ \left\lfloor \frac{10}{7} \right\rfloor = 1 $

So the prime factorization is:

$$
10! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7^1
$$

---

### Step 2: Use the Structure of $ x^3 \mid 10! $

We need $ x^3 $ to be a **factor** of $ 10! $, meaning that:

$$
x^3 \mid 10! \Rightarrow \text{each exponent of prime in } x^3 \leq \text{corresponding exponent in } 10!
$$

Let the prime factorization of $ x $ be:

$$
x = 2^a \cdot 3^b \cdot 5^c \cdot 7^d
$$

Then:

$$
x^3 = 2^{3a} \cdot 3^{3b} \cdot 5^{3c} \cdot 7^{3d}
$$

Now, impose the divisibility condition:

- $ 3a \leq 8 $ ⇒ $ a \leq \left\lfloor \frac{8}{3} \right\rfloor = 2 $
- $ 3b \leq 4 $ ⇒ $ b \leq \left\lfloor \frac{4}{3} \right\rfloor = 1 $
- $ 3c \leq 2 $ ⇒ $ c \leq \left\lfloor \frac{2}{3} \right\rfloor = 0 $
- $ 3d \leq 1 $ ⇒ $ d \leq \left\lfloor \frac{1}{3} \right\rfloor = 0 $

So, the possible exponents are:

- $ a \in \{0, 1, 2\} $ → 3 choices
- $ b \in \{0, 1\} $ → 2 choices
- $ c = 0 $, $ d = 0 $ → 1 choice each

---

### Step 3: Calculate Total Number of Valid $ x $

Each valid exponent combination gives a unique positive integer value of $ x $. The total number of values is:

$$
3 \times 2 \times 1 \times 1 = 6
$$

---

### Step 4: List All Valid $ x $ Values

Let’s enumerate them:

- $ a = 0, b = 0 $ ⇒ $ x = 1 $
- $ a = 0, b = 1 $ ⇒ $ x = 3 $
- $ a = 1, b = 0 $ ⇒ $ x = 2 $
- $ a = 1, b = 1 $ ⇒ $ x = 6 $
- $ a = 2, b = 0 $ ⇒ $ x = 4 $
- $ a = 2, b = 1 $ ⇒ $ x = 12 $

These values satisfy the condition that $ x^3 \mid 10! $. Any additional values would either involve primes not in $ 10! $ (like 5 or 7) or exponents that violate the divisibility condition.

---

### Final Answer

$$
\boxed{6}
$$

Solution:
First, we factor $10!:$
\begin{align*} 10!&=10\cdot 9\cdot 8\cdot 7 \cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\\ &=2^8\cdot 3^4 \cdot 5^2 \cdot 7.\end{align*}
Hence, $x$ can be $1, 2^1, 2^2, 3, 2^1\cdot 3,\text{ or }2^2\cdot 3$ for a total of $\boxed{6}$ possible values of $x.$