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, since the expression involves both x and y, maybe I can use calculus and find critical points by taking partial derivatives? Or maybe there's a way to simplify it by substitution? Let me try both approaches and see which one works.

Let me first consider substitution. Maybe I can set variables such that u = x and v = y, but that doesn't really help. Alternatively, maybe I can set t = x - y? Wait, but the denominator is x⁴ + y⁴ + 6. Not sure if that helps directly.

Alternatively, maybe I can assume some symmetry? For example, if I set y = -x or something? Let me try that. Suppose y = -x. Then the numerator becomes x - (-x) = 2x. The denominator becomes x⁴ + (-x)⁴ + 6 = 2x⁴ + 6. So the expression becomes 2x/(2x⁴ + 6) = x/(x⁴ + 3). Then, maybe I can find the maximum of this function in terms of x. Let me try that.

Let me define f(x) = x/(x⁴ + 3). To find its maximum, take derivative f’(x) = [1*(x⁴ + 3) - x*(4x³)]/(x⁴ + 3)^2 = [x⁴ + 3 - 4x⁴]/(denominator) = [ -3x⁴ + 3 ] / (denominator). Setting numerator equal to zero: -3x⁴ + 3 = 0 => x⁴ = 1 => x = ±1.

So critical points at x = 1 and x = -1. Let me check f(1) = 1/(1 + 3) = 1/4. f(-1) = -1/(1 + 3) = -1/4. So maximum in this case is 1/4. But is this the maximum over all real numbers x and y? Maybe not. Because I assumed y = -x. Maybe there's a higher value when y is not equal to -x.

Alternatively, maybe I can use substitution variables. Let me think. Let me set variables such that u = x + y and v = x - y? Not sure. Alternatively, maybe set variables to make the denominator symmetric? Let me think. The denominator is x⁴ + y⁴ + 6. Since x⁴ and y⁴ are both even functions, maybe the maximum occurs when x and y have some relation?

Alternatively, maybe use the method of Lagrange multipliers? Let me try that. Let me consider the function f(x, y) = (x - y)/(x⁴ + y⁴ + 6). To find its maximum, we can set the gradient equal to zero.

So compute partial derivatives. Let me denote D = x⁴ + y⁴ + 6. Then f = (x - y)/D.

Partial derivative with respect to x: [1*D - (x - y)*(4x³)] / D². Similarly, partial derivative with respect to y: [-1*D - (x - y)*(4y³)] / D².

Set both partial derivatives to zero. So:

For ∂f/∂x = 0:

D - (x - y)(4x³) = 0

Similarly, ∂f/∂y = 0:

-D - (x - y)(4y³) = 0

So we have two equations:

1) D - 4x³(x - y) = 0

2) -D - 4y³(x - y) = 0

Let me write them again:

Equation (1): x⁴ + y⁴ + 6 - 4x³(x - y) = 0

Equation (2): - (x⁴ + y⁴ + 6) - 4y³(x - y) = 0

Let me denote S = x - y. Then, equation (1) becomes:

x⁴ + y⁴ + 6 - 4x³ S = 0

Equation (2) becomes:

- (x⁴ + y⁴ + 6) - 4y³ S = 0

Let me add equations (1) and (2):

[ x⁴ + y⁴ + 6 - 4x³ S ] + [ -x⁴ - y⁴ - 6 - 4y³ S ] = 0 + 0

Simplify:

x⁴ - x⁴ + y⁴ - y⁴ + 6 - 6 -4x³ S -4y³ S = 0 => -4S(x³ + y³) = 0

So either S = 0 or x³ + y³ = 0.

Case 1: S = 0. That is, x - y = 0 => x = y. Then, substituting back into equation (1):

x⁴ + x⁴ + 6 - 4x³(0) = 0 => 2x⁴ + 6 = 0. But 2x⁴ + 6 is always positive, so no solution here. Therefore, Case 1 gives no critical points.

Case 2: x³ + y³ = 0. Which implies y³ = -x³ => y = -x (since real numbers). So again, y = -x. Wait, this is the same substitution as before. So then, substituting y = -x into equation (1):

x⁴ + (-x)^4 + 6 - 4x³(x - (-x)) = 0 => 2x⁴ + 6 - 4x³(2x) = 0 => 2x⁴ + 6 - 8x⁴ = 0 => -6x⁴ + 6 = 0 => x⁴ = 1 => x = ±1. Which gives us the same critical points as before. Therefore, the maximum in this case is 1/4.

But wait, so according to Lagrange multipliers, the only critical points are when y = -x, leading to maximum 1/4. But is this the global maximum? Maybe there are other critical points? Wait, but we considered all possibilities from the Lagrange multipliers. Since when we added equations (1) and (2), we found that either S = 0 or x³ + y³ = 0. Since S = 0 gives no solution, only x³ + y³ = 0. Therefore, the only critical points are when y = -x. Therefore, the maximum is 1/4? But maybe I need to check other possibilities?

Wait, but maybe there are other critical points where the denominator is zero? Wait, denominator is x⁴ + y⁴ + 6, which is always positive, since x⁴ and y⁴ are non-negative, so the minimum value is 6. Therefore, denominator is always positive, so the function is defined for all real x and y. Therefore, the critical points are only those found above. Therefore, the maximum value is 1/4? But let me check with some other values.

For example, suppose x is very large, say x approaches infinity, and y is fixed. Then numerator is x - y ~ x, denominator is x⁴ + ... which tends to infinity. Therefore, the whole expression tends to zero. Similarly, if x approaches negative infinity, numerator tends to negative infinity, denominator tends to infinity, so expression tends to zero. Similarly for y. So maybe the maximum is indeed achieved at some finite point.

But let me check another case. Suppose y = 0. Then the expression becomes x/(x⁴ + 0 + 6) = x/(x⁴ + 6). Let me find its maximum. Take derivative: [1*(x⁴ + 6) - x*(4x³)]/(x⁴ + 6)^2 = [x⁴ + 6 - 4x⁴]/denominator = (-3x⁴ + 6)/denominator. Setting numerator to zero: -3x⁴ + 6 = 0 => x⁴ = 2 => x = ±√[4]{2}. Then, f(x) = x/(x⁴ + 6) = ±√[4]{2}/(2 + 6) = ±√[4]{2}/8. Since √[4]{2} is 2^{1/4} ≈ 1.189, so positive value is approx 1.189/8 ≈ 0.1486, which is less than 1/4 = 0.25. So indeed, less than the previous case.

Similarly, if x = 0, then expression becomes (-y)/(0 + y⁴ + 6). Let me set y positive. Then, it's -y/(y⁴ + 6). The maximum of this would be negative, so not helpful for maximum.

Alternatively, suppose we set y = kx, where k is a constant. Maybe this substitution can help. Let me try that. Let me set y = kx. Then, the expression becomes (x - kx)/(x⁴ + (k x)^4 + 6) = x(1 - k)/[x⁴(1 + k⁴) + 6]. Let me denote this as f(x) = [x(1 - k)] / [x⁴(1 + k⁴) + 6]. To find maximum over x.

Let me consider for fixed k. Let me set t = x. Then f(t) = [t(1 - k)] / [t⁴(1 + k⁴) + 6]. To find maximum of this function. Let me take derivative with respect to t:

Numerator derivative: [ (1 - k)(t⁴(1 + k⁴) + 6) - t(1 - k)(4t³(1 + k⁴)) ] / [denominator]^2

Set numerator equal to zero:

(1 - k)(t⁴(1 + k⁴) + 6) - t(1 - k)(4t³(1 + k⁴)) = 0

Assuming 1 - k ≠ 0, we can divide both sides by (1 - k):

t⁴(1 + k⁴) + 6 - 4t⁴(1 + k⁴) = 0 => -3t⁴(1 + k⁴) + 6 = 0 => t⁴ = 6 / [3(1 + k⁴)] = 2 / (1 + k⁴)

Therefore, t = ± [2/(1 + k⁴)]^{1/4}

Therefore, the maximum value for fixed k is:

[ t(1 - k) ] / [ t⁴(1 + k⁴) + 6 ]

But t⁴ = 2/(1 + k⁴), so denominator becomes [2/(1 + k⁴)]*(1 + k⁴) + 6 = 2 + 6 = 8.

Numerator is t(1 - k). Since t = ± [2/(1 + k⁴)]^{1/4}, so the maximum value (taking positive t if 1 - k is positive) would be [ (1 - k) * [2/(1 + k⁴)]^{1/4} ] / 8.

Therefore, the maximum value for fixed k is (1 - k)/8 * [2/(1 + k⁴)]^{1/4}

So now, we need to maximize this expression over k. Let me denote this as:

f(k) = (1 - k)/8 * [2/(1 + k⁴)]^{1/4}

This seems complicated, but maybe we can find its maximum. Let me set m = k. Let me try to find maximum of f(k).

Alternatively, maybe set k = 1? Then f(1) = 0. Not helpful. If k = 0, then f(0) = (1 - 0)/8 * [2/(1 + 0)]^{1/4} = 1/8 * 2^{1/4} ≈ 1/8 * 1.189 ≈ 0.1486, which is same as before.

If k approaches infinity, then 1 - k approaches -infty, and [2/(1 + k⁴)]^{1/4} approaches zero. So the product might approach zero? Not sure. Let me check. Let me write f(k) as:

(1 - k) * [2]^{1/4} / [8*(1 + k⁴)^{1/4}]

So as k approaches infinity, numerator ~ -k, denominator ~ 8*k, so overall ~ -k/(8k) = -1/8. So approaches -1/8. So negative.

If k approaches -infty, then 1 - k approaches +infty, and [2/(1 + k⁴)]^{1/4} ~ [2/k⁴]^{1/4} = 2^{1/4}/k. So f(k) ~ (1 - k) * 2^{1/4}/k / 8 ~ (-k) * 2^{1/4}/k / 8 = -2^{1/4}/8 ≈ -0.1486.

So maybe maximum occurs somewhere finite. Let me try k = -1. Then f(-1) = (1 - (-1))/8 * [2/(1 + 1)]^{1/4} = 2/8 * [2/2]^{1/4} = (1/4)*1 = 1/4. Oh! That's the same value as before. So when k = -1, which corresponds to y = -x, we get 1/4.

So maybe this is the maximum? Let me check another value. Let me try k = something else. For example, k = 1/2. Then f(k) = (1 - 1/2)/8 * [2/(1 + (1/2)^4)]^{1/4} = (1/2)/8 * [2/(1 + 1/16)]^{1/4} = (1/16) * [2/(17/16)]^{1/4} = (1/16) * [32/17]^{1/4}. Let me compute 32/17 ≈ 1.882, so 1.882^{1/4} ≈ 1.16. Then 1/16 * 1.16 ≈ 0.0725, which is less than 1/4.

How about k = something else? Let me try k = -2. Then f(-2) = (1 - (-2))/8 * [2/(1 + (-2)^4)]^{1/4} = 3/8 * [2/(1 + 16)]^{1/4} = 3/8 * [2/17]^{1/4} ≈ 3/8 * 0.47 ≈ 0.176, which is still less than 1/4.

How about k = -0.5? Then f(-0.5) = (1 - (-0.5))/8 * [2/(1 + (-0.5)^4)]^{1/4} = 1.5/8 * [2/(1 + 0.0625)]^{1/4} = 1.5/8 * [2/1.0625]^{1/4} ≈ 0.1875 * [1.882]^{1/4} ≈ 0.1875 * 1.16 ≈ 0.217, which is still less than 1/4.

So maybe the maximum is indeed 1/4? But let me check another approach. Let me consider variables substitution. Let me set u = x^4 + y^4 + 6. But not sure. Alternatively, maybe use inequalities.

Since the denominator is x⁴ + y⁴ + 6, and the numerator is x - y. Maybe I can bound the numerator and denominator.

Note that by AM-GM inequality, x⁴ + y⁴ ≥ 2x²y². But not sure if helpful. Alternatively, maybe use Cauchy-Schwarz?

Alternatively, think of x - y as a linear term and denominator as quartic. Maybe use substitution variables. Let me think. Let me set t = x - y. Then, I need to express the denominator in terms of t. But how?

Alternatively, maybe fix t = x - y and try to minimize the denominator? For fixed t, find the minimum of x⁴ + y⁴ + 6. But since x and y are related by t = x - y, maybe express y = x - t. Then, substitute into denominator: x⁴ + (x - t)^4 + 6. Then, for fixed t, find the minimum of this expression over x. Then, the original expression would be t / [min(x⁴ + (x - t)^4 + 6)]. Then, the maximum of this over t would be the answer.

But this seems complicated. Let me try. Let me denote f(x) = x⁴ + (x - t)^4. To find its minimum. Take derivative: f’(x) = 4x³ + 4(x - t)^3. Set to zero: 4x³ + 4(x - t)^3 = 0 => x³ + (x - t)^3 = 0. Let me expand (x - t)^3: x³ - 3x²t + 3xt² - t³. Therefore, x³ + x³ - 3x²t + 3xt² - t³ = 0 => 2x³ - 3x²t + 3xt² - t³ = 0.

This seems complicated. Maybe factor? Let me try to factor. Let me set z = x. Then equation is 2z³ - 3t z² + 3t² z - t³ = 0. Let me try to factor this. Maybe factor as (z - a)(quadratic). Let me try z = t/2? Let me plug z = t/2:

2*(t/2)^3 - 3t*(t/2)^2 + 3t²*(t/2) - t³ = 2*(t³/8) - 3t*(t²/4) + 3t³/2 - t³ = t³/4 - 3t³/4 + 3t³/2 - t³ = (1/4 - 3/4 + 3/2 - 1)t³ = (-1/2 + 3/2 - 1)t³ = (1 - 1)t³ = 0. So z = t/2 is a root. Therefore, we can factor (z - t/2) from the cubic.

Let me perform polynomial division or use synthetic division. Let me write the cubic as 2z³ - 3t z² + 3t² z - t³. Divided by (z - t/2). Let me use polynomial division.

Alternatively, write as 2z³ - 3t z² + 3t² z - t³ = (z - t/2)(Az² + Bz + C). Multiply out RHS: z*(Az² + Bz + C) - t/2*(Az² + Bz + C) = Az³ + Bz² + Cz - (A t / 2) z² - (B t / 2) z - (C t / 2).

Set equal to original polynomial:

Az³ + Bz² + Cz - (A t / 2) z² - (B t / 2) z - (C t / 2) = 2z³ - 3t z² + 3t² z - t³.

Therefore, equate coefficients:

For z³: A = 2.

For z²: B - (A t / 2) = -3t => B - (2t / 2) = -3t => B - t = -3t => B = -2t.

For z: C - (B t / 2) = 3t². Since B = -2t, then C - (-2t * t / 2) = C + t² = 3t² => C = 2t².

For constant term: -C t / 2 = -t³ => - (2t² * t)/2 = -t³ => -t³ = -t³. Correct.

Therefore, the cubic factors as (z - t/2)(2z² - 2t z + 2t²). Therefore, the equation 2z³ - 3t z² + 3t² z - t³ = 0 factors as (z - t/2)(2z² - 2t z + 2t²) = 0.

Therefore, solutions are z = t/2 and solutions of 2z² - 2t z + 2t² = 0. Let me solve 2z² - 2t z + 2t² = 0. Divide by 2: z² - t z + t² = 0. Discriminant: t² - 4t² = -3t² < 0. Therefore, only real solution is z = t/2. Therefore, the minimum occurs at x = t/2.

Therefore, for fixed t, the minimum of x⁴ + (x - t)^4 occurs at x = t/2. Therefore, substituting back, x = t/2, y = x - t = t/2 - t = -t/2. Therefore, the minimum of the denominator is (t/2)^4 + (-t/2)^4 + 6 = 2*(t^4)/16 + 6 = t^4 / 8 + 6.

Therefore, the original expression (x - y)/denominator becomes t / (t^4 / 8 + 6) = 8t / (t^4 + 48). Therefore, now the problem reduces to maximizing 8t / (t^4 + 48) over real numbers t.

So now, we need to find maximum of f(t) = 8t / (t⁴ + 48). Let me compute its maximum.

Take derivative: f’(t) = [8(t⁴ + 48) - 8t*(4t³)] / (t⁴ + 48)^2 = [8t⁴ + 384 - 32t⁴]/(denominator) = [ -24t⁴ + 384 ] / (denominator). Setting numerator equal to zero: -24t⁴ + 384 = 0 => t⁴ = 384 / 24 = 16 => t = ±2.

Therefore, critical points at t = 2 and t = -2.

Compute f(2) = 8*2 / (16 + 48) = 16 / 64 = 1/4.

Compute f(-2) = 8*(-2)/(16 + 48) = -16/64 = -1/4.

Therefore, maximum value is 1/4.

So this approach also leads to maximum value 1/4. Therefore, this seems to confirm the previous result.

But wait, earlier when I considered y = -x, which corresponds to t = x - y = x - (-x) = 2x. So if t = 2, then x = 1, y = -1. Which is the same as before. Similarly, t = -2 would be x = -1, y = 1.

Therefore, regardless of the approach, the maximum value is 1/4.

But let me check if there are other possibilities. For example, suppose I don't assume y = kx or t = x - y. Maybe there's another case where the maximum is higher?

Alternatively, maybe consider using the method of substitution with variables. Let me think. Let me set u = x^2 and v = y^2. But since x and y can be positive or negative, maybe not helpful.

Alternatively, note that x^4 + y^4 ≥ (x² + y²)^2 / 2 by Cauchy-Schwarz? Wait, actually, (x^4 + y^4)(1 + 1) ≥ (x² + y²)^2 by Cauchy-Schwarz. Therefore, x^4 + y^4 ≥ (x² + y²)^2 / 2. But not sure if helpful here.

Alternatively, maybe use substitution variables. Let me set a = x^2, b = y^2. Then, since x and y can be positive or negative, but the numerator is x - y. However, this complicates things because of the sign.

Alternatively, think of x - y as a linear term and try to bound it. For example, using Hölder's inequality?

But maybe not necessary, since we already have two methods leading to 1/4. Also, testing other points didn't give higher values.

Alternatively, let me check if 1/4 is indeed the maximum. Suppose I take x = 1, y = -1. Then numerator is 1 - (-1) = 2. Denominator is 1 + 1 + 6 = 8. So 2/8 = 1/4. If I take x = 1, y = -1 + ε, then numerator becomes 1 - (-1 + ε) = 2 - ε. Denominator becomes 1 + (-1 + ε)^4 + 6. Let me compute for small ε. Let me take ε approaching 0. Then denominator is approximately 1 + 1 + 6 = 8. So the expression is approximately (2 - ε)/8, which is slightly less than 1/4. Similarly, if I take x slightly larger than 1, y slightly larger than -1, maybe? Let me try x = 1 + δ, y = -1 + δ. Then numerator is (1 + δ) - (-1 + δ) = 2. Denominator is (1 + δ)^4 + (-1 + δ)^4 + 6. Let me compute for small δ. Expand (1 + δ)^4 ≈ 1 + 4δ + 6δ² + 4δ³ + δ⁴. Similarly, (-1 + δ)^4 = (1 - δ)^4 ≈ 1 - 4δ + 6δ² - 4δ³ + δ⁴. Adding them together: 2 + 12δ² + 2δ⁴. Therefore, denominator ≈ 2 + 12δ² + 2δ⁴ + 6 = 8 + 12δ². Therefore, the expression is 2 / (8 + 12δ²) ≈ 2/8 * [1 - (12δ²)/8 + ...] ≈ 1/4 - (3δ²)/4 + ... which is less than 1/4. Therefore, perturbing around x = 1, y = -1 gives lower value.

Therefore, this suggests that 1/4 is indeed the maximum.

Alternatively, let me check another point. Suppose x = √2, y = -√2. Then numerator is √2 - (-√2) = 2√2 ≈ 2.828. Denominator is (√2)^4 + (-√2)^4 + 6 = 4 + 4 + 6 = 14. So the expression is 2√2 / 14 ≈ 0.202, which is less than 1/4 (which is 0.25).

Another test: x = 2, y = -2. Numerator is 4. Denominator is 16 + 16 + 6 = 38. 4/38 ≈ 0.105, less than 1/4.

How about x = 1. Let me try x = 1, y = something else. Suppose y = 0. Then as before, value is 1/7 ≈ 0.142, less than 1/4.

Alternatively, suppose x = 1, y = -1 + t. Let me take t = 0.5. Then y = -0.5. Numerator is 1 - (-0.5) = 1.5. Denominator is 1 + (-0.5)^4 + 6 = 1 + 0.0625 + 6 = 7.0625. So 1.5 / 7.0625 ≈ 0.212, still less than 0.25.

Alternatively, take x = 1. Let me try to maximize (1 - y)/(1 + y⁴ + 6) = (1 - y)/(y⁴ + 7). Take derivative with respect to y: [ -1*(y⁴ + 7) - (1 - y)(4y³) ] / (denominator)^2. Set numerator to zero: -y⁴ -7 -4y³ + 4y⁴ = 0 => 3y⁴ -4y³ -7 = 0. This seems complicated. Maybe try y = -1: 3(1) -4(-1)^3 -7 = 3 + 4 -7 = 0. So y = -1 is a root. Therefore, factor out (y + 1). Let me perform polynomial division.

Divide 3y⁴ -4y³ -7 by (y + 1). Using synthetic division:

Coefficients: 3, -4, 0, 0, -7 (since original polynomial is 3y⁴ -4y³ + 0y² + 0y -7)

Root at y = -1:

Bring down 3. Multiply by -1: -3. Add to next coefficient: -4 + (-3) = -7. Multiply by -1: 7. Add to next coefficient: 0 + 7 = 7. Multiply by -1: -7. Add to next coefficient: 0 + (-7) = -7. Multiply by -1: 7. Add to last coefficient: -7 + 7 = 0.

Therefore, polynomial factors as (y + 1)(3y³ -7y² +7y -7). Now, solve 3y³ -7y² +7y -7 = 0. Try rational roots. Possible roots are ±1, ±7, ±1/3, ±7/3. Testing y = 1: 3 -7 +7 -7 = -4 ≠ 0. y = 7: too big. y = 7/3: messy. Maybe no real roots? Let me check derivative of 3y³ -7y² +7y -7. Derivative is 9y² -14y +7. Discriminant: 196 - 252 = -56 < 0. Therefore, the cubic has no real roots. Therefore, only real root is y = -1. Therefore, maximum occurs at y = -1, which gives (1 - (-1))/(1 + 1 + 6) = 2/8 = 1/4. So again, same result.

Therefore, all approaches lead to the conclusion that the maximum value is 1/4.

But just to be thorough, let me check if there are any other critical points when we considered Lagrange multipliers. Earlier, we saw that the only critical points are when y = -x, leading to maximum 1/4. Therefore, unless there are other critical points where the gradient is undefined, but since the function is smooth everywhere, there are no other critical points.

Therefore, I think it's safe to conclude that the maximum value 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 by analyzing the structure of the function and applying optimization techniques.

---

### Step 1: Understanding the Function

The function is a rational expression where the **numerator** is linear in $ x $ and $ y $, and the **denominator** is a sum of even powers of $ x $ and $ y $, plus a constant. Since $ x^4 $ and $ y^4 $ are always non-negative, the denominator is always **positive**, ensuring the function is defined for all real $ x $ and $ y $.

---

### Step 2: Try a Symmetric Substitution

Let’s consider the substitution $ y = -x $. Then:

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

So the expression becomes:

$$
\frac{2x}{2x^4 + 6} = \frac{x}{x^4 + 3}
$$

Now define:

$$
f(x) = \frac{x}{x^4 + 3}
$$

To find the maximum of this function, take the derivative:

$$
f'(x) = \frac{(x^4 + 3) - x(4x^3)}{(x^4 + 3)^2} = \frac{-3x^4 + 3}{(x^4 + 3)^2}
$$

Set the numerator to zero:

$$
-3x^4 + 3 = 0 \Rightarrow x^4 = 1 \Rightarrow x = \pm 1
$$

Evaluate $ f(x) $ at these points:

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

So the **maximum** value in this case is $ \frac{1}{4} $.

---

### Step 3: General Case via Lagrange Multipliers

Let’s define the function:

$$
f(x, y) = \frac{x - y}{x^4 + y^4 + 6}
$$

Let $ D = x^4 + y^4 + 6 $, so $ f(x, y) = \frac{x - y}{D} $

Compute partial derivatives:

- $ \frac{\partial f}{\partial x} = \frac{D - (x - y)(4x^3)}{D^2} $
- $ \frac{\partial f}{\partial y} = \frac{-D - (x - y)(4y^3)}{D^2} $

Set both partial derivatives to zero:

1. $ D - 4x^3(x - y) = 0 $
2. $ -D - 4y^3(x - y) = 0 $

Adding the two equations:

$$
-4(x - y)(x^3 + y^3) = 0
$$

So either $ x = y $ or $ x^3 + y^3 = 0 $

- If $ x = y $, then the numerator becomes 0, so the function is 0.
- If $ x^3 + y^3 = 0 $, then $ y = -x $, which is the same substitution as before.

Thus, the only critical points occur when $ y = -x $, and the maximum value is again $ \frac{1}{4} $.

---

### Step 4: Confirm with a General Substitution

Let’s define $ t = x - y $, and express $ y = x - t $. Then the expression becomes:

$$
\frac{t}{x^4 + (x - t)^4 + 6}
$$

Minimizing the denominator for fixed $ t $ leads to the minimum of $ x^4 + (x - t)^4 $, which occurs at $ x = \frac{t}{2} $, $ y = -\frac{t}{2} $

So the expression becomes:

$$
\frac{t}{\left(\frac{t}{2}\right)^4 + \left(-\frac{t}{2}\right)^4 + 6} = \frac{t}{\frac{t^4}{8} + 6} = \frac{8t}{t^4 + 48}
$$

Now define:

$$
g(t) = \frac{8t}{t^4 + 48}
$$

Take derivative:

$$
g'(t) = \frac{8(t^4 + 48) - 8t(4t^3)}{(t^4 + 48)^2} = \frac{-24t^4 + 384}{(t^4 + 48)^2}
$$

Set numerator to zero:

$$
-24t^4 + 384 = 0 \Rightarrow t^4 = 16 \Rightarrow t = \pm 2
$$

Evaluate:

- $ g(2) = \frac{16}{16 + 48} = \frac{16}{64} = \frac{1}{4} $
- $ g(-2) = \frac{-16}{16 + 48} = -\frac{1}{4} $

So the **maximum** value is again $ \frac{1}{4} $.

---

### Step 5: Conclusion

All approaches—substitution, Lagrange multipliers, and direct optimization—lead to the same conclusion: the **maximum value** of the expression is achieved when $ x = 1 $, $ y = -1 $, or $ x = -1 $, $ y = 1 $, and the value 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, since the expression involves both x and y, maybe I can use calculus and find critical points by taking partial derivatives? Or maybe there's a way to simplify it by substitution? Let me try both approaches and see which one works.

Let me first consider substitution. Maybe I can set variables such that u = x and v = y, but that doesn't really help. Alternatively, maybe I can set t = x - y? Wait, but the denominator is x⁴ + y⁴ + 6. Not sure if that helps directly.

Alternatively, maybe I can assume some symmetry? For example, if I set y = -x or something? Let me try that. Suppose y = -x. Then the numerator becomes x - (-x) = 2x. The denominator becomes x⁴ + (-x)⁴ + 6 = 2x⁴ + 6. So the expression becomes 2x/(2x⁴ + 6) = x/(x⁴ + 3). Then, maybe I can find the maximum of this function in terms of x. Let me try that.

Let me define f(x) = x/(x⁴ + 3). To find its maximum, take derivative f’(x) = [1*(x⁴ + 3) - x*(4x³)]/(x⁴ + 3)^2 = [x⁴ + 3 - 4x⁴]/(denominator) = [ -3x⁴ + 3 ] / (denominator). Setting numerator equal to zero: -3x⁴ + 3 = 0 => x⁴ = 1 => x = ±1.

So critical points at x = 1 and x = -1. Let me check f(1) = 1/(1 + 3) = 1/4. f(-1) = -1/(1 + 3) = -1/4. So maximum in this case is 1/4. But is this the maximum over all real numbers x and y? Maybe not. Because I assumed y = -x. Maybe there's a higher value when y is not equal to -x.

Alternatively, maybe I can use substitution variables. Let me think. Let me set variables such that u = x + y and v = x - y? Not sure. Alternatively, maybe set variables to make the denominator symmetric? Let me think. The denominator is x⁴ + y⁴ + 6. Since x⁴ and y⁴ are both even functions, maybe the maximum occurs when x and y have some relation?

Alternatively, maybe use the method of Lagrange multipliers? Let me try that. Let me consider the function f(x, y) = (x - y)/(x⁴ + y⁴ + 6). To find its maximum, we can set the gradient equal to zero.

So compute partial derivatives. Let me denote D = x⁴ + y⁴ + 6. Then f = (x - y)/D.

Partial derivative with respect to x: [1*D - (x - y)*(4x³)] / D². Similarly, partial derivative with respect to y: [-1*D - (x - y)*(4y³)] / D².

Set both partial derivatives to zero. So:

For ∂f/∂x = 0:

D - (x - y)(4x³) = 0

Similarly, ∂f/∂y = 0:

-D - (x - y)(4y³) = 0

So we have two equations:

1) D - 4x³(x - y) = 0

2) -D - 4y³(x - y) = 0

Let me write them again:

Equation (1): x⁴ + y⁴ + 6 - 4x³(x - y) = 0

Equation (2): - (x⁴ + y⁴ + 6) - 4y³(x - y) = 0

Let me denote S = x - y. Then, equation (1) becomes:

x⁴ + y⁴ + 6 - 4x³ S = 0

Equation (2) becomes:

- (x⁴ + y⁴ + 6) - 4y³ S = 0

Let me add equations (1) and (2):

[ x⁴ + y⁴ + 6 - 4x³ S ] + [ -x⁴ - y⁴ - 6 - 4y³ S ] = 0 + 0

Simplify:

x⁴ - x⁴ + y⁴ - y⁴ + 6 - 6 -4x³ S -4y³ S = 0 => -4S(x³ + y³) = 0

So either S = 0 or x³ + y³ = 0.

Case 1: S = 0. That is, x - y = 0 => x = y. Then, substituting back into equation (1):

x⁴ + x⁴ + 6 - 4x³(0) = 0 => 2x⁴ + 6 = 0. But 2x⁴ + 6 is always positive, so no solution here. Therefore, Case 1 gives no critical points.

Case 2: x³ + y³ = 0. Which implies y³ = -x³ => y = -x (since real numbers). So again, y = -x. Wait, this is the same substitution as before. So then, substituting y = -x into equation (1):

x⁴ + (-x)^4 + 6 - 4x³(x - (-x)) = 0 => 2x⁴ + 6 - 4x³(2x) = 0 => 2x⁴ + 6 - 8x⁴ = 0 => -6x⁴ + 6 = 0 => x⁴ = 1 => x = ±1. Which gives us the same critical points as before. Therefore, the maximum in this case is 1/4.

But wait, so according to Lagrange multipliers, the only critical points are when y = -x, leading to maximum 1/4. But is this the global maximum? Maybe there are other critical points? Wait, but we considered all possibilities from the Lagrange multipliers. Since when we added equations (1) and (2), we found that either S = 0 or x³ + y³ = 0. Since S = 0 gives no solution, only x³ + y³ = 0. Therefore, the only critical points are when y = -x. Therefore, the maximum is 1/4? But maybe I need to check other possibilities?

Wait, but maybe there are other critical points where the denominator is zero? Wait, denominator is x⁴ + y⁴ + 6, which is always positive, since x⁴ and y⁴ are non-negative, so the minimum value is 6. Therefore, denominator is always positive, so the function is defined for all real x and y. Therefore, the critical points are only those found above. Therefore, the maximum value is 1/4? But let me check with some other values.

For example, suppose x is very large, say x approaches infinity, and y is fixed. Then numerator is x - y ~ x, denominator is x⁴ + ... which tends to infinity. Therefore, the whole expression tends to zero. Similarly, if x approaches negative infinity, numerator tends to negative infinity, denominator tends to infinity, so expression tends to zero. Similarly for y. So maybe the maximum is indeed achieved at some finite point.

But let me check another case. Suppose y = 0. Then the expression becomes x/(x⁴ + 0 + 6) = x/(x⁴ + 6). Let me find its maximum. Take derivative: [1*(x⁴ + 6) - x*(4x³)]/(x⁴ + 6)^2 = [x⁴ + 6 - 4x⁴]/denominator = (-3x⁴ + 6)/denominator. Setting numerator to zero: -3x⁴ + 6 = 0 => x⁴ = 2 => x = ±√[4]{2}. Then, f(x) = x/(x⁴ + 6) = ±√[4]{2}/(2 + 6) = ±√[4]{2}/8. Since √[4]{2} is 2^{1/4} ≈ 1.189, so positive value is approx 1.189/8 ≈ 0.1486, which is less than 1/4 = 0.25. So indeed, less than the previous case.

Similarly, if x = 0, then expression becomes (-y)/(0 + y⁴ + 6). Let me set y positive. Then, it's -y/(y⁴ + 6). The maximum of this would be negative, so not helpful for maximum.

Alternatively, suppose we set y = kx, where k is a constant. Maybe this substitution can help. Let me try that. Let me set y = kx. Then, the expression becomes (x - kx)/(x⁴ + (k x)^4 + 6) = x(1 - k)/[x⁴(1 + k⁴) + 6]. Let me denote this as f(x) = [x(1 - k)] / [x⁴(1 + k⁴) + 6]. To find maximum over x.

Let me consider for fixed k. Let me set t = x. Then f(t) = [t(1 - k)] / [t⁴(1 + k⁴) + 6]. To find maximum of this function. Let me take derivative with respect to t:

Numerator derivative: [ (1 - k)(t⁴(1 + k⁴) + 6) - t(1 - k)(4t³(1 + k⁴)) ] / [denominator]^2

Set numerator equal to zero:

(1 - k)(t⁴(1 + k⁴) + 6) - t(1 - k)(4t³(1 + k⁴)) = 0

Assuming 1 - k ≠ 0, we can divide both sides by (1 - k):

t⁴(1 + k⁴) + 6 - 4t⁴(1 + k⁴) = 0 => -3t⁴(1 + k⁴) + 6 = 0 => t⁴ = 6 / [3(1 + k⁴)] = 2 / (1 + k⁴)

Therefore, t = ± [2/(1 + k⁴)]^{1/4}

Therefore, the maximum value for fixed k is:

[ t(1 - k) ] / [ t⁴(1 + k⁴) + 6 ]

But t⁴ = 2/(1 + k⁴), so denominator becomes [2/(1 + k⁴)]*(1 + k⁴) + 6 = 2 + 6 = 8.

Numerator is t(1 - k). Since t = ± [2/(1 + k⁴)]^{1/4}, so the maximum value (taking positive t if 1 - k is positive) would be [ (1 - k) * [2/(1 + k⁴)]^{1/4} ] / 8.

Therefore, the maximum value for fixed k is (1 - k)/8 * [2/(1 + k⁴)]^{1/4}

So now, we need to maximize this expression over k. Let me denote this as:

f(k) = (1 - k)/8 * [2/(1 + k⁴)]^{1/4}

This seems complicated, but maybe we can find its maximum. Let me set m = k. Let me try to find maximum of f(k).

Alternatively, maybe set k = 1? Then f(1) = 0. Not helpful. If k = 0, then f(0) = (1 - 0)/8 * [2/(1 + 0)]^{1/4} = 1/8 * 2^{1/4} ≈ 1/8 * 1.189 ≈ 0.1486, which is same as before.

If k approaches infinity, then 1 - k approaches -infty, and [2/(1 + k⁴)]^{1/4} approaches zero. So the product might approach zero? Not sure. Let me check. Let me write f(k) as:

(1 - k) * [2]^{1/4} / [8*(1 + k⁴)^{1/4}]

So as k approaches infinity, numerator ~ -k, denominator ~ 8*k, so overall ~ -k/(8k) = -1/8. So approaches -1/8. So negative.

If k approaches -infty, then 1 - k approaches +infty, and [2/(1 + k⁴)]^{1/4} ~ [2/k⁴]^{1/4} = 2^{1/4}/k. So f(k) ~ (1 - k) * 2^{1/4}/k / 8 ~ (-k) * 2^{1/4}/k / 8 = -2^{1/4}/8 ≈ -0.1486.

So maybe maximum occurs somewhere finite. Let me try k = -1. Then f(-1) = (1 - (-1))/8 * [2/(1 + 1)]^{1/4} = 2/8 * [2/2]^{1/4} = (1/4)*1 = 1/4. Oh! That's the same value as before. So when k = -1, which corresponds to y = -x, we get 1/4.

So maybe this is the maximum? Let me check another value. Let me try k = something else. For example, k = 1/2. Then f(k) = (1 - 1/2)/8 * [2/(1 + (1/2)^4)]^{1/4} = (1/2)/8 * [2/(1 + 1/16)]^{1/4} = (1/16) * [2/(17/16)]^{1/4} = (1/16) * [32/17]^{1/4}. Let me compute 32/17 ≈ 1.882, so 1.882^{1/4} ≈ 1.16. Then 1/16 * 1.16 ≈ 0.0725, which is less than 1/4.

How about k = something else? Let me try k = -2. Then f(-2) = (1 - (-2))/8 * [2/(1 + (-2)^4)]^{1/4} = 3/8 * [2/(1 + 16)]^{1/4} = 3/8 * [2/17]^{1/4} ≈ 3/8 * 0.47 ≈ 0.176, which is still less than 1/4.

How about k = -0.5? Then f(-0.5) = (1 - (-0.5))/8 * [2/(1 + (-0.5)^4)]^{1/4} = 1.5/8 * [2/(1 + 0.0625)]^{1/4} = 1.5/8 * [2/1.0625]^{1/4} ≈ 0.1875 * [1.882]^{1/4} ≈ 0.1875 * 1.16 ≈ 0.217, which is still less than 1/4.

So maybe the maximum is indeed 1/4? But let me check another approach. Let me consider variables substitution. Let me set u = x^4 + y^4 + 6. But not sure. Alternatively, maybe use inequalities.

Since the denominator is x⁴ + y⁴ + 6, and the numerator is x - y. Maybe I can bound the numerator and denominator.

Note that by AM-GM inequality, x⁴ + y⁴ ≥ 2x²y². But not sure if helpful. Alternatively, maybe use Cauchy-Schwarz?

Alternatively, think of x - y as a linear term and denominator as quartic. Maybe use substitution variables. Let me think. Let me set t = x - y. Then, I need to express the denominator in terms of t. But how?

Alternatively, maybe fix t = x - y and try to minimize the denominator? For fixed t, find the minimum of x⁴ + y⁴ + 6. But since x and y are related by t = x - y, maybe express y = x - t. Then, substitute into denominator: x⁴ + (x - t)^4 + 6. Then, for fixed t, find the minimum of this expression over x. Then, the original expression would be t / [min(x⁴ + (x - t)^4 + 6)]. Then, the maximum of this over t would be the answer.

But this seems complicated. Let me try. Let me denote f(x) = x⁴ + (x - t)^4. To find its minimum. Take derivative: f’(x) = 4x³ + 4(x - t)^3. Set to zero: 4x³ + 4(x - t)^3 = 0 => x³ + (x - t)^3 = 0. Let me expand (x - t)^3: x³ - 3x²t + 3xt² - t³. Therefore, x³ + x³ - 3x²t + 3xt² - t³ = 0 => 2x³ - 3x²t + 3xt² - t³ = 0.

This seems complicated. Maybe factor? Let me try to factor. Let me set z = x. Then equation is 2z³ - 3t z² + 3t² z - t³ = 0. Let me try to factor this. Maybe factor as (z - a)(quadratic). Let me try z = t/2? Let me plug z = t/2:

2*(t/2)^3 - 3t*(t/2)^2 + 3t²*(t/2) - t³ = 2*(t³/8) - 3t*(t²/4) + 3t³/2 - t³ = t³/4 - 3t³/4 + 3t³/2 - t³ = (1/4 - 3/4 + 3/2 - 1)t³ = (-1/2 + 3/2 - 1)t³ = (1 - 1)t³ = 0. So z = t/2 is a root. Therefore, we can factor (z - t/2) from the cubic.

Let me perform polynomial division or use synthetic division. Let me write the cubic as 2z³ - 3t z² + 3t² z - t³. Divided by (z - t/2). Let me use polynomial division.

Alternatively, write as 2z³ - 3t z² + 3t² z - t³ = (z - t/2)(Az² + Bz + C). Multiply out RHS: z*(Az² + Bz + C) - t/2*(Az² + Bz + C) = Az³ + Bz² + Cz - (A t / 2) z² - (B t / 2) z - (C t / 2).

Set equal to original polynomial:

Az³ + Bz² + Cz - (A t / 2) z² - (B t / 2) z - (C t / 2) = 2z³ - 3t z² + 3t² z - t³.

Therefore, equate coefficients:

For z³: A = 2.

For z²: B - (A t / 2) = -3t => B - (2t / 2) = -3t => B - t = -3t => B = -2t.

For z: C - (B t / 2) = 3t². Since B = -2t, then C - (-2t * t / 2) = C + t² = 3t² => C = 2t².

For constant term: -C t / 2 = -t³ => - (2t² * t)/2 = -t³ => -t³ = -t³. Correct.

Therefore, the cubic factors as (z - t/2)(2z² - 2t z + 2t²). Therefore, the equation 2z³ - 3t z² + 3t² z - t³ = 0 factors as (z - t/2)(2z² - 2t z + 2t²) = 0.

Therefore, solutions are z = t/2 and solutions of 2z² - 2t z + 2t² = 0. Let me solve 2z² - 2t z + 2t² = 0. Divide by 2: z² - t z + t² = 0. Discriminant: t² - 4t² = -3t² < 0. Therefore, only real solution is z = t/2. Therefore, the minimum occurs at x = t/2.

Therefore, for fixed t, the minimum of x⁴ + (x - t)^4 occurs at x = t/2. Therefore, substituting back, x = t/2, y = x - t = t/2 - t = -t/2. Therefore, the minimum of the denominator is (t/2)^4 + (-t/2)^4 + 6 = 2*(t^4)/16 + 6 = t^4 / 8 + 6.

Therefore, the original expression (x - y)/denominator becomes t / (t^4 / 8 + 6) = 8t / (t^4 + 48). Therefore, now the problem reduces to maximizing 8t / (t^4 + 48) over real numbers t.

So now, we need to find maximum of f(t) = 8t / (t⁴ + 48). Let me compute its maximum.

Take derivative: f’(t) = [8(t⁴ + 48) - 8t*(4t³)] / (t⁴ + 48)^2 = [8t⁴ + 384 - 32t⁴]/(denominator) = [ -24t⁴ + 384 ] / (denominator). Setting numerator equal to zero: -24t⁴ + 384 = 0 => t⁴ = 384 / 24 = 16 => t = ±2.

Therefore, critical points at t = 2 and t = -2.

Compute f(2) = 8*2 / (16 + 48) = 16 / 64 = 1/4.

Compute f(-2) = 8*(-2)/(16 + 48) = -16/64 = -1/4.

Therefore, maximum value is 1/4.

So this approach also leads to maximum value 1/4. Therefore, this seems to confirm the previous result.

But wait, earlier when I considered y = -x, which corresponds to t = x - y = x - (-x) = 2x. So if t = 2, then x = 1, y = -1. Which is the same as before. Similarly, t = -2 would be x = -1, y = 1.

Therefore, regardless of the approach, the maximum value is 1/4.

But let me check if there are other possibilities. For example, suppose I don't assume y = kx or t = x - y. Maybe there's another case where the maximum is higher?

Alternatively, maybe consider using the method of substitution with variables. Let me think. Let me set u = x^2 and v = y^2. But since x and y can be positive or negative, maybe not helpful.

Alternatively, note that x^4 + y^4 ≥ (x² + y²)^2 / 2 by Cauchy-Schwarz? Wait, actually, (x^4 + y^4)(1 + 1) ≥ (x² + y²)^2 by Cauchy-Schwarz. Therefore, x^4 + y^4 ≥ (x² + y²)^2 / 2. But not sure if helpful here.

Alternatively, maybe use substitution variables. Let me set a = x^2, b = y^2. Then, since x and y can be positive or negative, but the numerator is x - y. However, this complicates things because of the sign.

Alternatively, think of x - y as a linear term and try to bound it. For example, using Hölder's inequality?

But maybe not necessary, since we already have two methods leading to 1/4. Also, testing other points didn't give higher values.

Alternatively, let me check if 1/4 is indeed the maximum. Suppose I take x = 1, y = -1. Then numerator is 1 - (-1) = 2. Denominator is 1 + 1 + 6 = 8. So 2/8 = 1/4. If I take x = 1, y = -1 + ε, then numerator becomes 1 - (-1 + ε) = 2 - ε. Denominator becomes 1 + (-1 + ε)^4 + 6. Let me compute for small ε. Let me take ε approaching 0. Then denominator is approximately 1 + 1 + 6 = 8. So the expression is approximately (2 - ε)/8, which is slightly less than 1/4. Similarly, if I take x slightly larger than 1, y slightly larger than -1, maybe? Let me try x = 1 + δ, y = -1 + δ. Then numerator is (1 + δ) - (-1 + δ) = 2. Denominator is (1 + δ)^4 + (-1 + δ)^4 + 6. Let me compute for small δ. Expand (1 + δ)^4 ≈ 1 + 4δ + 6δ² + 4δ³ + δ⁴. Similarly, (-1 + δ)^4 = (1 - δ)^4 ≈ 1 - 4δ + 6δ² - 4δ³ + δ⁴. Adding them together: 2 + 12δ² + 2δ⁴. Therefore, denominator ≈ 2 + 12δ² + 2δ⁴ + 6 = 8 + 12δ². Therefore, the expression is 2 / (8 + 12δ²) ≈ 2/8 * [1 - (12δ²)/8 + ...] ≈ 1/4 - (3δ²)/4 + ... which is less than 1/4. Therefore, perturbing around x = 1, y = -1 gives lower value.

Therefore, this suggests that 1/4 is indeed the maximum.

Alternatively, let me check another point. Suppose x = √2, y = -√2. Then numerator is √2 - (-√2) = 2√2 ≈ 2.828. Denominator is (√2)^4 + (-√2)^4 + 6 = 4 + 4 + 6 = 14. So the expression is 2√2 / 14 ≈ 0.202, which is less than 1/4 (which is 0.25).

Another test: x = 2, y = -2. Numerator is 4. Denominator is 16 + 16 + 6 = 38. 4/38 ≈ 0.105, less than 1/4.

How about x = 1. Let me try x = 1, y = something else. Suppose y = 0. Then as before, value is 1/7 ≈ 0.142, less than 1/4.

Alternatively, suppose x = 1, y = -1 + t. Let me take t = 0.5. Then y = -0.5. Numerator is 1 - (-0.5) = 1.5. Denominator is 1 + (-0.5)^4 + 6 = 1 + 0.0625 + 6 = 7.0625. So 1.5 / 7.0625 ≈ 0.212, still less than 0.25.

Alternatively, take x = 1. Let me try to maximize (1 - y)/(1 + y⁴ + 6) = (1 - y)/(y⁴ + 7). Take derivative with respect to y: [ -1*(y⁴ + 7) - (1 - y)(4y³) ] / (denominator)^2. Set numerator to zero: -y⁴ -7 -4y³ + 4y⁴ = 0 => 3y⁴ -4y³ -7 = 0. This seems complicated. Maybe try y = -1: 3(1) -4(-1)^3 -7 = 3 + 4 -7 = 0. So y = -1 is a root. Therefore, factor out (y + 1). Let me perform polynomial division.

Divide 3y⁴ -4y³ -7 by (y + 1). Using synthetic division:

Coefficients: 3, -4, 0, 0, -7 (since original polynomial is 3y⁴ -4y³ + 0y² + 0y -7)

Root at y = -1:

Bring down 3. Multiply by -1: -3. Add to next coefficient: -4 + (-3) = -7. Multiply by -1: 7. Add to next coefficient: 0 + 7 = 7. Multiply by -1: -7. Add to next coefficient: 0 + (-7) = -7. Multiply by -1: 7. Add to last coefficient: -7 + 7 = 0.

Therefore, polynomial factors as (y + 1)(3y³ -7y² +7y -7). Now, solve 3y³ -7y² +7y -7 = 0. Try rational roots. Possible roots are ±1, ±7, ±1/3, ±7/3. Testing y = 1: 3 -7 +7 -7 = -4 ≠ 0. y = 7: too big. y = 7/3: messy. Maybe no real roots? Let me check derivative of 3y³ -7y² +7y -7. Derivative is 9y² -14y +7. Discriminant: 196 - 252 = -56 < 0. Therefore, the cubic has no real roots. Therefore, only real root is y = -1. Therefore, maximum occurs at y = -1, which gives (1 - (-1))/(1 + 1 + 6) = 2/8 = 1/4. So again, same result.

Therefore, all approaches lead to the conclusion that the maximum value is 1/4.

But just to be thorough, let me check if there are any other critical points when we considered Lagrange multipliers. Earlier, we saw that the only critical points are when y = -x, leading to maximum 1/4. Therefore, unless there are other critical points where the gradient is undefined, but since the function is smooth everywhere, there are no other critical points.

Therefore, I think it's safe to conclude that the maximum value 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 by analyzing the structure of the function and applying optimization techniques.

---

### Step 1: Understanding the Function

The function is a rational expression where the **numerator** is linear in $ x $ and $ y $, and the **denominator** is a sum of even powers of $ x $ and $ y $, plus a constant. Since $ x^4 $ and $ y^4 $ are always non-negative, the denominator is always **positive**, ensuring the function is defined for all real $ x $ and $ y $.

---

### Step 2: Try a Symmetric Substitution

Let’s consider the substitution $ y = -x $. Then:

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

So the expression becomes:

$$
\frac{2x}{2x^4 + 6} = \frac{x}{x^4 + 3}
$$

Now define:

$$
f(x) = \frac{x}{x^4 + 3}
$$

To find the maximum of this function, take the derivative:

$$
f'(x) = \frac{(x^4 + 3) - x(4x^3)}{(x^4 + 3)^2} = \frac{-3x^4 + 3}{(x^4 + 3)^2}
$$

Set the numerator to zero:

$$
-3x^4 + 3 = 0 \Rightarrow x^4 = 1 \Rightarrow x = \pm 1
$$

Evaluate $ f(x) $ at these points:

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

So the **maximum** value in this case is $ \frac{1}{4} $.

---

### Step 3: General Case via Lagrange Multipliers

Let’s define the function:

$$
f(x, y) = \frac{x - y}{x^4 + y^4 + 6}
$$

Let $ D = x^4 + y^4 + 6 $, so $ f(x, y) = \frac{x - y}{D} $

Compute partial derivatives:

- $ \frac{\partial f}{\partial x} = \frac{D - (x - y)(4x^3)}{D^2} $
- $ \frac{\partial f}{\partial y} = \frac{-D - (x - y)(4y^3)}{D^2} $

Set both partial derivatives to zero:

1. $ D - 4x^3(x - y) = 0 $
2. $ -D - 4y^3(x - y) = 0 $

Adding the two equations:

$$
-4(x - y)(x^3 + y^3) = 0
$$

So either $ x = y $ or $ x^3 + y^3 = 0 $

- If $ x = y $, then the numerator becomes 0, so the function is 0.
- If $ x^3 + y^3 = 0 $, then $ y = -x $, which is the same substitution as before.

Thus, the only critical points occur when $ y = -x $, and the maximum value is again $ \frac{1}{4} $.

---

### Step 4: Confirm with a General Substitution

Let’s define $ t = x - y $, and express $ y = x - t $. Then the expression becomes:

$$
\frac{t}{x^4 + (x - t)^4 + 6}
$$

Minimizing the denominator for fixed $ t $ leads to the minimum of $ x^4 + (x - t)^4 $, which occurs at $ x = \frac{t}{2} $, $ y = -\frac{t}{2} $

So the expression becomes:

$$
\frac{t}{\left(\frac{t}{2}\right)^4 + \left(-\frac{t}{2}\right)^4 + 6} = \frac{t}{\frac{t^4}{8} + 6} = \frac{8t}{t^4 + 48}
$$

Now define:

$$
g(t) = \frac{8t}{t^4 + 48}
$$

Take derivative:

$$
g'(t) = \frac{8(t^4 + 48) - 8t(4t^3)}{(t^4 + 48)^2} = \frac{-24t^4 + 384}{(t^4 + 48)^2}
$$

Set numerator to zero:

$$
-24t^4 + 384 = 0 \Rightarrow t^4 = 16 \Rightarrow t = \pm 2
$$

Evaluate:

- $ g(2) = \frac{16}{16 + 48} = \frac{16}{64} = \frac{1}{4} $
- $ g(-2) = \frac{-16}{16 + 48} = -\frac{1}{4} $

So the **maximum** value is again $ \frac{1}{4} $.

---

### Step 5: Conclusion

All approaches—substitution, Lagrange multipliers, and direct optimization—lead to the same conclusion: the **maximum value** of the expression is achieved when $ x = 1 $, $ y = -1 $, or $ x = -1 $, $ y = 1 $, and the value 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. 10 factorial is the product of all integers from 1 to 10, so 10! = 10 × 9 × 8 × ... × 1. Let me compute that value first, but maybe I don't need the exact number. Instead, I think I need to look at the prime factorization of 10! because factors are related to prime exponents.

Right, so if I can find the prime factors of 10! and their exponents, then I can figure out what exponents x³ can have. Since 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!.

Let me recall how to find the prime factorization of 10!. The primes less than or equal to 10 are 2, 3, 5, 7. Let me find the exponents of each prime in 10!.

Starting with 2: The exponent of 2 in 10! is given by the sum of floor(10/2) + floor(10/4) + floor(10/8) + ... until the division results in zero. Let me compute that:

floor(10/2) = 5

floor(10/4) = 2

floor(10/8) = 1

floor(10/16) = 0, so we stop here. So total exponent of 2 is 5 + 2 + 1 = 8.

Similarly for 3:

floor(10/3) = 3

floor(10/9) = 1

floor(10/27) = 0. So total exponent of 3 is 3 + 1 = 4.

For 5:

floor(10/5) = 2

floor(10/25) = 0. So exponent of 5 is 2.

For 7:

floor(10/7) = 1

floor(10/49) = 0. So exponent of 7 is 1.

So the prime factorization of 10! is 2⁸ × 3⁴ × 5² × 7¹.

Now, if x³ divides 10!, then x must be an integer such that when you take its prime factorization, each exponent is at most the exponent in 10! divided by 3, right? Wait, no. Let me think again.

Suppose x has a prime factorization of x = 2^a × 3^b × 5^c × 7^d × ... Then x³ would be 2^(3a) × 3^(3b) × 5^(3c) × 7^(3d) × ... For x³ to divide 10!, each exponent in x³ must be less than or equal to the corresponding exponent in 10!.

Therefore, for each prime:

3a ≤ 8 (for prime 2)

3b ≤ 4 (for prime 3)

3c ≤ 2 (for prime 5)

3d ≤ 1 (for prime 7)

And for any other primes, since 10! doesn't have them, their exponents would have to be zero. So x can't have any prime factors other than 2, 3, 5, 7.

Therefore, the exponents a, b, c, d must satisfy:

For 2: 3a ≤ 8 ⇒ a ≤ 8/3 ≈ 2.666... So since a must be an integer, a ≤ 2.

Similarly for 3: 3b ≤ 4 ⇒ b ≤ 4/3 ≈ 1.333... ⇒ b ≤ 1.

For 5: 3c ≤ 2 ⇒ c ≤ 2/3 ≈ 0.666... ⇒ c ≤ 0. Since c must be a non-negative integer, c = 0.

For 7: 3d ≤ 1 ⇒ d ≤ 1/3 ≈ 0.333... ⇒ d = 0.

Therefore, the exponents for each prime in x are:

a: 0, 1, or 2 (since a ≤ 2)

b: 0 or 1 (since b ≤ 1)

c: 0 (since c must be 0)

d: 0 (since d must be 0)

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. Is that all? Let me check again.

Wait, but x is an integer. So x can be 1, 2, 3, 4, 6, or something else? Let me list them.

Wait, if a can be 0,1,2 and b can be 0,1, then:

Possible combinations:

a=0, b=0: x=2^0 * 3^0 = 1

a=0, b=1: x=3^1 = 3

a=1, b=0: x=2^1 = 2

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

a=2, b=0: x=2²=4

a=2, b=1: x=2²*3=12

So that's 6 values: 1, 3, 2, 6, 4, 12. So 6 values. But wait, is 12³ a factor of 10!?

Let me check. Let me compute 12³. 12³ = (2²×3)^3 = 2^6 × 3^3. Now, in 10! we have 2^8 and 3^4. So 2^6 ≤ 2^8 and 3^3 ≤ 3^4. So yes, 12³ divides 10!.

Similarly, check 6³: 6³ = (2×3)^3 = 2³×3³. Which is okay since 2³ ≤ 2^8 and 3³ ≤ 3^4. So yes.

Similarly, 4³ = 2^6, which is okay. 3³ is okay. 2³ is okay. 1³ is okay. So all 6 values are valid. So answer is 6?

But wait, the question says "how many possible integer values of x are there?" So 6? But let me think again. Wait, maybe I missed something?

Wait, but maybe there are more possibilities? Because maybe x can have other primes? But no, because 10! doesn't have any other primes. So x can't have primes other than 2, 3, 5, 7. But for 5 and 7, their exponents in x must be zero. Because if you have even one 5 in x, then x³ would have 3c ≥ 3*1 = 3, but in 10! the exponent of 5 is only 2, so 3c ≤ 2 ⇒ c=0. Similarly for 7, 3d ≤ 1 ⇒ d=0. Therefore, x cannot have 5 or 7 as factors. Therefore, x can only be composed of 2s and 3s. Therefore, the number of x is indeed 3*2=6. So answer is 6?

But wait, let me check again. Let me think if there are other numbers. For example, is 1 considered? Yes, 1³ is 1, which divides anything. So 1 is included. So 1, 2, 3, 4, 6, 12. That's 6 numbers. So answer is 6? But I have a feeling that maybe I need to check if there are more numbers. Wait, maybe I made a mistake in the exponents?

Let me check again the exponents for each prime.

For prime 2: exponent in 10! is 8. So 3a ≤ 8 ⇒ a ≤ 2.666, so a can be 0,1,2. Correct.

For prime 3: exponent in 10! is 4. So 3b ≤ 4 ⇒ b ≤ 1.333, so b can be 0 or 1. Correct.

For prime 5: exponent in 10! is 2. So 3c ≤ 2 ⇒ c ≤ 0.666 ⇒ c=0. Correct.

For prime 7: exponent in 10! is 1. So 3d ≤ 1 ⇒ d=0. Correct.

Therefore, the exponents for 2 and 3 are as above. Therefore, the number of possible x is 3*2=6. Therefore, answer is 6? But let me check if there are other numbers. For example, is 2*3* something? Wait, but we already considered that. For example, 2*3=6, which is included. What about numbers like 2^2*3^1=12, which is included. What about numbers with higher exponents? For example, if a=2, b=1, that's 12. If a=2, b=0, that's 4. If a=1, b=1, that's 6. If a=1, b=0, that's 2. If a=0, b=1, that's 3. If a=0, b=0, that's 1. So that's all. So 6 numbers. Therefore, answer is 6?

But wait, I remember that sometimes when you have exponents, you have to consider that exponents can be zero. But in this case, we already considered that. So maybe 6 is correct. But let me check with another approach.

Alternatively, think of x³ divides 10! So x divides 10!^(1/3). But since 10! is not a perfect cube, but we can think of the exponents. Wait, but maybe not helpful.

Alternatively, think of the exponents of primes in x. For each prime, the exponent in x is floor(exponent in 10! / 3). Wait, no. Wait, if x³ divides 10!, then for each prime p, exponent of p in x³ is ≤ exponent of p in 10!. Therefore, exponent of p in x is ≤ floor(exponent of p in 10! / 3). Wait, no. Wait, if exponent of p in x is k, then exponent of p in x³ is 3k. Therefore, 3k ≤ exponent of p in 10!.

Therefore, k ≤ floor(exponent of p in 10! / 3). Wait, but floor(exponent / 3) is not necessarily the same as floor(exponent / 3). Wait, for example, for prime 2, exponent is 8. So 8 / 3 is 2.666..., so floor(8/3) is 2. Which is same as what we had before. Similarly for prime 3: exponent is 4, 4/3 is 1.333..., floor is 1. For prime 5: exponent is 2, 2/3 is 0.666..., floor is 0. For prime 7: exponent is 1, floor(1/3)=0. Therefore, the exponents for each prime in x are floor(exponent / 3). Therefore, the number of possible exponents for each prime is floor(exponent / 3) + 1. Wait, because exponents can be from 0 to floor(exponent / 3). So for prime 2: floor(8/3)=2, so exponents 0,1,2: 3 choices. For prime 3: floor(4/3)=1, exponents 0,1: 2 choices. For prime 5: floor(2/3)=0, exponents 0: 1 choice. For prime 7: floor(1/3)=0, exponents 0: 1 choice. Therefore, total number of x is 3*2*1*1=6. So same result. Therefore, answer is 6. Therefore, the number of possible integer values of x is 6. Therefore, the answer is \boxed{6}?

Wait, but let me check again. Let me think if there are other numbers. For example, suppose x is 1, 2, 3, 4, 6, 12. Are there any other numbers?

Wait, what about 2^2 * 3^0 = 4, which is included. 2^1 * 3^1 = 6. 2^0 * 3^1 = 3. 2^2 * 3^1 = 12. 2^1 * 3^0 = 2. 2^0 * 3^0 = 1. So that's all. So 6 numbers. So answer is 6. But I have a doubt because sometimes problems like this can have more answers. Let me check if maybe exponents can be higher?

Wait, for example, suppose someone thinks that maybe exponents can be higher if they are not multiples? Wait, no. Because if you have, say, for prime 2, exponent in x is 2, then exponent in x³ is 6, which is less than 8. If you tried exponent 3, then 3*3=9, which is more than 8. So that's not allowed. Similarly for prime 3, exponent in x can be 1, because 3*1=3 ≤ 4. If you tried 2, then 3*2=6 > 4. So not allowed. For prime 5, exponent in x can only be 0. For prime 7, same. Therefore, no, there are no other exponents. Therefore, answer is 6. Therefore, I think the answer is 6. But let me check with actual numbers.

Let me list all possible x:

1. x=1: 1³=1 divides 10! Yes.

2. x=2: 2³=8 divides 10! Yes, since 10! has 2^8.

3. x=3: 3³=27 divides 10! Yes, since 10! has 3^4.

4. x=4: 4³=64. 64 is 2^6. 10! has 2^8, so yes.

5. x=6: 6³=216. 216 is 2³×3³. 10! has 2^8 and 3^4, so yes.

6. x=12: 12³=1728. 1728 is 2^6×3³. Which is okay, as above.

Are there any other numbers? Let's think. What about x= 2^2 * 3^1 = 12, which is already included. What about x= 2^1 * 3^1 = 6, included. What about x= 2^0 * 3^1 = 3, included. What about x= 2^2 * 3^0 = 4, included. What about x= 2^1 * 3^0 = 2, included. What about x= 2^0 * 3^0 = 1, included. So that's all. So 6 numbers. Therefore, answer is 6. Therefore, I think the answer is 6. But let me check if maybe there are numbers with exponents beyond what I considered? For example, if someone thought that maybe exponents can be fractions? But no, exponents have to be integers. So no. Therefore, I think the answer is 6. Wait, but I remember that sometimes when you have multiple primes, you can have combinations. But in this case, since 5 and 7 can't be in x, there are no other primes. Therefore, only combinations of 2 and 3. Therefore, 3*2=6. So answer is 6. Therefore, I think the answer is 6. But let me check once more.

Alternatively, maybe I can think of the problem as follows: Find the number of positive integers x such that x³ divides 10!.

This is equivalent to finding the number of positive integers x such that x divides (10!)^{1/3}. But since 10! is not a perfect cube, this is not straightforward. However, we can think in terms of exponents. For each prime in the factorization of x, the exponent must satisfy 3k ≤ exponent in 10!.

Which is what I did before. Therefore, the number of such x is the product over each prime of (floor(exponent / 3) + 1). Wait, no. Wait, for each prime, the number of choices for the exponent is floor(exponent / 3) + 1? Wait, no. Wait, if the exponent in 10! is e, then the exponent in x can be from 0 up to floor(e / 3). Therefore, the number of choices is floor(e / 3) + 1. Wait, for example, for prime 2, e=8, floor(8/3)=2, so number of choices is 2 + 1 = 3. Which matches. For prime 3, e=4, floor(4/3)=1, so 1 + 1 = 2. For prime 5, e=2, floor(2/3)=0, so 0 + 1 = 1. For prime 7, e=1, floor(1/3)=0, so 0 + 1 = 1. Therefore, total number of x is 3*2*1*1=6. Therefore, same result. Therefore, answer is 6. Therefore, I think the answer is 6. Therefore, I think the answer is \boxed{6}.

But just to be thorough, let me check if there are any other numbers. For example, suppose someone considers x= 2^2 * 3^1 * 5^0 * 7^0 = 12, which is included. What about x= 2^2 * 3^1 * 5^0 * 7^0 = 12. What about x= 2^2 * 3^1 * 5^0 * 7^0? That's same as 12. What about x= 2^2 * 3^0 * 5^0 * 7^0 = 4. So that's included. What about x= 2^1 * 3^1 * 5^0 * 7^0 = 6. Included. What about x= 2^0 * 3^1 * 5^0 * 7^0 = 3. Included. x= 2^1 * 3^0 * 5^0 * 7^0 = 2. Included. x= 1. So that's all. Therefore, 6 numbers. Therefore, answer is 6. Therefore, I think the answer is 6. Therefore, I think the answer is \boxed{6}.

**Final Answer**
\boxed{6}

To determine how many possible integer values of $ x $ satisfy the condition that $ x^3 $ is a positive factor of $ 10! $, we begin by analyzing the prime factorization of $ 10! $.

---

### Step 1: Prime Factorization of $ 10! $

We compute the exponents of the prime factors in $ 10! $ using Legendre's formula:

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

- **Prime 3**:
  $$
  \left\lfloor \frac{10}{3} \right\rfloor + \left\lfloor \frac{10}{9} \right\rfloor = 3 + 1 = 4
  $$

- **Prime 5**:
  $$
  \left\lfloor \frac{10}{5} \right\rfloor = 2
  $$

- **Prime 7**:
  $$
  \left\lfloor \frac{10}{7} \right\rfloor = 1
  $$

So, the prime factorization of $ 10! $ is:
$$
10! = 2^8 \times 3^4 \times 5^2 \times 7^1
$$

---

### Step 2: Conditions for $ x^3 \mid 10! $

Let $ x = 2^a \times 3^b \times 5^c \times 7^d $, where $ a, b, c, d \in \mathbb{Z}_{\geq 0} $. Then:
$$
x^3 = 2^{3a} \times 3^{3b} \times 5^{3c} \times 7^{3d}
$$

For $ x^3 $ to divide $ 10! $, we must have:
$$
3a \leq 8, \quad 3b \leq 4, \quad 3c \leq 2, \quad 3d \leq 1
$$

Solving these inequalities:

- $ a \leq \left\lfloor \frac{8}{3} \right\rfloor = 2 $ → $ a \in \{0, 1, 2\} $ (3 choices)
- $ b \leq \left\lfloor \frac{4}{3} \right\rfloor = 1 $ → $ b \in \{0, 1\} $ (2 choices)
- $ c \leq \left\lfloor \frac{2}{3} \right\rfloor = 0 $ → $ c = 0 $ (1 choice)
- $ d \leq \left\lfloor \frac{1}{3} \right\rfloor = 0 $ → $ d = 0 $ (1 choice)

---

### Step 3: Counting the Number of Valid $ x $

Each valid combination of $ a, b, c, d $ corresponds to a unique integer $ x $. Since the choices for $ a $ and $ b $ are independent, the total number of such $ x $ is:
$$
3 \times 2 \times 1 \times 1 = 6
$$

---

### Step 4: Listing the Valid $ x $ Values

We can list the possible values of $ x $ by considering all combinations of $ a $ and $ b $:

- $ 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 are all the valid values of $ x $ such that $ x^3 \mid 10! $.

---

### 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. 10 factorial is the product of all integers from 1 to 10, so 10! = 10 × 9 × 8 × ... × 1. Let me compute that value first, but maybe I don't need the exact number. Instead, I think I need to look at the prime factorization of 10! because factors are related to prime exponents.

Right, so if I can find the prime factors of 10! and their exponents, then I can figure out what exponents x³ can have. Since 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!.

Let me recall how to find the prime factorization of 10!. The primes less than or equal to 10 are 2, 3, 5, 7. Let me find the exponents of each prime in 10!.

Starting with 2: The exponent of 2 in 10! is given by the sum of floor(10/2) + floor(10/4) + floor(10/8) + ... until the division results in zero. Let me compute that:

floor(10/2) = 5

floor(10/4) = 2

floor(10/8) = 1

floor(10/16) = 0, so we stop here. So total exponent of 2 is 5 + 2 + 1 = 8.

Similarly for 3:

floor(10/3) = 3

floor(10/9) = 1

floor(10/27) = 0. So total exponent of 3 is 3 + 1 = 4.

For 5:

floor(10/5) = 2

floor(10/25) = 0. So exponent of 5 is 2.

For 7:

floor(10/7) = 1

floor(10/49) = 0. So exponent of 7 is 1.

So the prime factorization of 10! is 2⁸ × 3⁴ × 5² × 7¹.

Now, if x³ divides 10!, then x must be an integer such that when you take its prime factorization, each exponent is at most the exponent in 10! divided by 3, right? Wait, no. Let me think again.

Suppose x has a prime factorization of x = 2^a × 3^b × 5^c × 7^d × ... Then x³ would be 2^(3a) × 3^(3b) × 5^(3c) × 7^(3d) × ... For x³ to divide 10!, each exponent in x³ must be less than or equal to the corresponding exponent in 10!.

Therefore, for each prime:

3a ≤ 8 (for prime 2)

3b ≤ 4 (for prime 3)

3c ≤ 2 (for prime 5)

3d ≤ 1 (for prime 7)

And for any other primes, since 10! doesn't have them, their exponents would have to be zero. So x can't have any prime factors other than 2, 3, 5, 7.

Therefore, the exponents a, b, c, d must satisfy:

For 2: 3a ≤ 8 ⇒ a ≤ 8/3 ≈ 2.666... So since a must be an integer, a ≤ 2.

Similarly for 3: 3b ≤ 4 ⇒ b ≤ 4/3 ≈ 1.333... ⇒ b ≤ 1.

For 5: 3c ≤ 2 ⇒ c ≤ 2/3 ≈ 0.666... ⇒ c ≤ 0. Since c must be a non-negative integer, c = 0.

For 7: 3d ≤ 1 ⇒ d ≤ 1/3 ≈ 0.333... ⇒ d = 0.

Therefore, the exponents for each prime in x are:

a: 0, 1, or 2 (since a ≤ 2)

b: 0 or 1 (since b ≤ 1)

c: 0 (since c must be 0)

d: 0 (since d must be 0)

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. Is that all? Let me check again.

Wait, but x is an integer. So x can be 1, 2, 3, 4, 6, or something else? Let me list them.

Wait, if a can be 0,1,2 and b can be 0,1, then:

Possible combinations:

a=0, b=0: x=2^0 * 3^0 = 1

a=0, b=1: x=3^1 = 3

a=1, b=0: x=2^1 = 2

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

a=2, b=0: x=2²=4

a=2, b=1: x=2²*3=12

So that's 6 values: 1, 3, 2, 6, 4, 12. So 6 values. But wait, is 12³ a factor of 10!?

Let me check. Let me compute 12³. 12³ = (2²×3)^3 = 2^6 × 3^3. Now, in 10! we have 2^8 and 3^4. So 2^6 ≤ 2^8 and 3^3 ≤ 3^4. So yes, 12³ divides 10!.

Similarly, check 6³: 6³ = (2×3)^3 = 2³×3³. Which is okay since 2³ ≤ 2^8 and 3³ ≤ 3^4. So yes.

Similarly, 4³ = 2^6, which is okay. 3³ is okay. 2³ is okay. 1³ is okay. So all 6 values are valid. So answer is 6?

But wait, the question says "how many possible integer values of x are there?" So 6? But let me think again. Wait, maybe I missed something?

Wait, but maybe there are more possibilities? Because maybe x can have other primes? But no, because 10! doesn't have any other primes. So x can't have primes other than 2, 3, 5, 7. But for 5 and 7, their exponents in x must be zero. Because if you have even one 5 in x, then x³ would have 3c ≥ 3*1 = 3, but in 10! the exponent of 5 is only 2, so 3c ≤ 2 ⇒ c=0. Similarly for 7, 3d ≤ 1 ⇒ d=0. Therefore, x cannot have 5 or 7 as factors. Therefore, x can only be composed of 2s and 3s. Therefore, the number of x is indeed 3*2=6. So answer is 6?

But wait, let me check again. Let me think if there are other numbers. For example, is 1 considered? Yes, 1³ is 1, which divides anything. So 1 is included. So 1, 2, 3, 4, 6, 12. That's 6 numbers. So answer is 6? But I have a feeling that maybe I need to check if there are more numbers. Wait, maybe I made a mistake in the exponents?

Let me check again the exponents for each prime.

For prime 2: exponent in 10! is 8. So 3a ≤ 8 ⇒ a ≤ 2.666, so a can be 0,1,2. Correct.

For prime 3: exponent in 10! is 4. So 3b ≤ 4 ⇒ b ≤ 1.333, so b can be 0 or 1. Correct.

For prime 5: exponent in 10! is 2. So 3c ≤ 2 ⇒ c ≤ 0.666 ⇒ c=0. Correct.

For prime 7: exponent in 10! is 1. So 3d ≤ 1 ⇒ d=0. Correct.

Therefore, the exponents for 2 and 3 are as above. Therefore, the number of possible x is 3*2=6. Therefore, answer is 6? But let me check if there are other numbers. For example, is 2*3* something? Wait, but we already considered that. For example, 2*3=6, which is included. What about numbers like 2^2*3^1=12, which is included. What about numbers with higher exponents? For example, if a=2, b=1, that's 12. If a=2, b=0, that's 4. If a=1, b=1, that's 6. If a=1, b=0, that's 2. If a=0, b=1, that's 3. If a=0, b=0, that's 1. So that's all. So 6 numbers. Therefore, answer is 6?

But wait, I remember that sometimes when you have exponents, you have to consider that exponents can be zero. But in this case, we already considered that. So maybe 6 is correct. But let me check with another approach.

Alternatively, think of x³ divides 10! So x divides 10!^(1/3). But since 10! is not a perfect cube, but we can think of the exponents. Wait, but maybe not helpful.

Alternatively, think of the exponents of primes in x. For each prime, the exponent in x is floor(exponent in 10! / 3). Wait, no. Wait, if x³ divides 10!, then for each prime p, exponent of p in x³ is ≤ exponent of p in 10!. Therefore, exponent of p in x is ≤ floor(exponent of p in 10! / 3). Wait, no. Wait, if exponent of p in x is k, then exponent of p in x³ is 3k. Therefore, 3k ≤ exponent of p in 10!.

Therefore, k ≤ floor(exponent of p in 10! / 3). Wait, but floor(exponent / 3) is not necessarily the same as floor(exponent / 3). Wait, for example, for prime 2, exponent is 8. So 8 / 3 is 2.666..., so floor(8/3) is 2. Which is same as what we had before. Similarly for prime 3: exponent is 4, 4/3 is 1.333..., floor is 1. For prime 5: exponent is 2, 2/3 is 0.666..., floor is 0. For prime 7: exponent is 1, floor(1/3)=0. Therefore, the exponents for each prime in x are floor(exponent / 3). Therefore, the number of possible exponents for each prime is floor(exponent / 3) + 1. Wait, because exponents can be from 0 to floor(exponent / 3). So for prime 2: floor(8/3)=2, so exponents 0,1,2: 3 choices. For prime 3: floor(4/3)=1, exponents 0,1: 2 choices. For prime 5: floor(2/3)=0, exponents 0: 1 choice. For prime 7: floor(1/3)=0, exponents 0: 1 choice. Therefore, total number of x is 3*2*1*1=6. So same result. Therefore, answer is 6. Therefore, the number of possible integer values of x is 6. Therefore, the answer is \boxed{6}?

Wait, but let me check again. Let me think if there are other numbers. For example, suppose x is 1, 2, 3, 4, 6, 12. Are there any other numbers?

Wait, what about 2^2 * 3^0 = 4, which is included. 2^1 * 3^1 = 6. 2^0 * 3^1 = 3. 2^2 * 3^1 = 12. 2^1 * 3^0 = 2. 2^0 * 3^0 = 1. So that's all. So 6 numbers. So answer is 6. But I have a doubt because sometimes problems like this can have more answers. Let me check if maybe exponents can be higher?

Wait, for example, suppose someone thinks that maybe exponents can be higher if they are not multiples? Wait, no. Because if you have, say, for prime 2, exponent in x is 2, then exponent in x³ is 6, which is less than 8. If you tried exponent 3, then 3*3=9, which is more than 8. So that's not allowed. Similarly for prime 3, exponent in x can be 1, because 3*1=3 ≤ 4. If you tried 2, then 3*2=6 > 4. So not allowed. For prime 5, exponent in x can only be 0. For prime 7, same. Therefore, no, there are no other exponents. Therefore, answer is 6. Therefore, I think the answer is 6. But let me check with actual numbers.

Let me list all possible x:

1. x=1: 1³=1 divides 10! Yes.

2. x=2: 2³=8 divides 10! Yes, since 10! has 2^8.

3. x=3: 3³=27 divides 10! Yes, since 10! has 3^4.

4. x=4: 4³=64. 64 is 2^6. 10! has 2^8, so yes.

5. x=6: 6³=216. 216 is 2³×3³. 10! has 2^8 and 3^4, so yes.

6. x=12: 12³=1728. 1728 is 2^6×3³. Which is okay, as above.

Are there any other numbers? Let's think. What about x= 2^2 * 3^1 = 12, which is already included. What about x= 2^1 * 3^1 = 6, included. What about x= 2^0 * 3^1 = 3, included. What about x= 2^2 * 3^0 = 4, included. What about x= 2^1 * 3^0 = 2, included. What about x= 2^0 * 3^0 = 1, included. So that's all. So 6 numbers. Therefore, answer is 6. Therefore, I think the answer is 6. But let me check if maybe there are numbers with exponents beyond what I considered? For example, if someone thought that maybe exponents can be fractions? But no, exponents have to be integers. So no. Therefore, I think the answer is 6. Wait, but I remember that sometimes when you have multiple primes, you can have combinations. But in this case, since 5 and 7 can't be in x, there are no other primes. Therefore, only combinations of 2 and 3. Therefore, 3*2=6. So answer is 6. Therefore, I think the answer is 6. But let me check once more.

Alternatively, maybe I can think of the problem as follows: Find the number of positive integers x such that x³ divides 10!.

This is equivalent to finding the number of positive integers x such that x divides (10!)^{1/3}. But since 10! is not a perfect cube, this is not straightforward. However, we can think in terms of exponents. For each prime in the factorization of x, the exponent must satisfy 3k ≤ exponent in 10!.

Which is what I did before. Therefore, the number of such x is the product over each prime of (floor(exponent / 3) + 1). Wait, no. Wait, for each prime, the number of choices for the exponent is floor(exponent / 3) + 1? Wait, no. Wait, if the exponent in 10! is e, then the exponent in x can be from 0 up to floor(e / 3). Therefore, the number of choices is floor(e / 3) + 1. Wait, for example, for prime 2, e=8, floor(8/3)=2, so number of choices is 2 + 1 = 3. Which matches. For prime 3, e=4, floor(4/3)=1, so 1 + 1 = 2. For prime 5, e=2, floor(2/3)=0, so 0 + 1 = 1. For prime 7, e=1, floor(1/3)=0, so 0 + 1 = 1. Therefore, total number of x is 3*2*1*1=6. Therefore, same result. Therefore, answer is 6. Therefore, I think the answer is 6. Therefore, I think the answer is \boxed{6}.

But just to be thorough, let me check if there are any other numbers. For example, suppose someone considers x= 2^2 * 3^1 * 5^0 * 7^0 = 12, which is included. What about x= 2^2 * 3^1 * 5^0 * 7^0 = 12. What about x= 2^2 * 3^1 * 5^0 * 7^0? That's same as 12. What about x= 2^2 * 3^0 * 5^0 * 7^0 = 4. So that's included. What about x= 2^1 * 3^1 * 5^0 * 7^0 = 6. Included. What about x= 2^0 * 3^1 * 5^0 * 7^0 = 3. Included. x= 2^1 * 3^0 * 5^0 * 7^0 = 2. Included. x= 1. So that's all. Therefore, 6 numbers. Therefore, answer is 6. Therefore, I think the answer is 6. Therefore, I think the answer is \boxed{6}.

**Final Answer**
\boxed{6}

To determine how many possible integer values of $ x $ satisfy the condition that $ x^3 $ is a positive factor of $ 10! $, we begin by analyzing the prime factorization of $ 10! $.

---

### Step 1: Prime Factorization of $ 10! $

We compute the exponents of the prime factors in $ 10! $ using Legendre's formula:

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

- **Prime 3**:
  $$
  \left\lfloor \frac{10}{3} \right\rfloor + \left\lfloor \frac{10}{9} \right\rfloor = 3 + 1 = 4
  $$

- **Prime 5**:
  $$
  \left\lfloor \frac{10}{5} \right\rfloor = 2
  $$

- **Prime 7**:
  $$
  \left\lfloor \frac{10}{7} \right\rfloor = 1
  $$

So, the prime factorization of $ 10! $ is:
$$
10! = 2^8 \times 3^4 \times 5^2 \times 7^1
$$

---

### Step 2: Conditions for $ x^3 \mid 10! $

Let $ x = 2^a \times 3^b \times 5^c \times 7^d $, where $ a, b, c, d \in \mathbb{Z}_{\geq 0} $. Then:
$$
x^3 = 2^{3a} \times 3^{3b} \times 5^{3c} \times 7^{3d}
$$

For $ x^3 $ to divide $ 10! $, we must have:
$$
3a \leq 8, \quad 3b \leq 4, \quad 3c \leq 2, \quad 3d \leq 1
$$

Solving these inequalities:

- $ a \leq \left\lfloor \frac{8}{3} \right\rfloor = 2 $ → $ a \in \{0, 1, 2\} $ (3 choices)
- $ b \leq \left\lfloor \frac{4}{3} \right\rfloor = 1 $ → $ b \in \{0, 1\} $ (2 choices)
- $ c \leq \left\lfloor \frac{2}{3} \right\rfloor = 0 $ → $ c = 0 $ (1 choice)
- $ d \leq \left\lfloor \frac{1}{3} \right\rfloor = 0 $ → $ d = 0 $ (1 choice)

---

### Step 3: Counting the Number of Valid $ x $

Each valid combination of $ a, b, c, d $ corresponds to a unique integer $ x $. Since the choices for $ a $ and $ b $ are independent, the total number of such $ x $ is:
$$
3 \times 2 \times 1 \times 1 = 6
$$

---

### Step 4: Listing the Valid $ x $ Values

We can list the possible values of $ x $ by considering all combinations of $ a $ and $ b $:

- $ 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 are all the valid values of $ x $ such that $ x^3 \mid 10! $.

---

### 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 $x.$