Explanation:
Okay, so Tom has this Mr. Potato Head, and he wants to figure out how many different wacky personalities he can make. Let me try to break this down step by step. First, let me list out all the parts he has:

- 3 hairstyles
- 2 sets of eyebrows
- 1 pair of googly eyes
- 2 sets of ears
- 2 sets of lips
- 1 pair of regular shoes
- 1 bonus pair of clown shoes

And the question says a complete personality includes eyebrows, eyes, ears, lips, shoes, and optionally hair. Also, note that you can't mix and match parts, like you can't take one eyebrow from one pair and another from another pair. So each set is considered as a whole, I guess? Or maybe each set has left and right? Wait, the note says you can't mix and match, so maybe each set is a pair? For example, the eyebrows are two sets, so maybe each set is a pair? Wait, but the problem says "2 sets of eyebrows". Hmm. Let me think.

Wait, the problem says "you cannot 'mix and match'. For example, you cannot take the left eyebrow from one pair and the right eyebrow from the other pair." So that suggests that each set of eyebrows is a pair, meaning that each set has a left and right eyebrow. So if you have two sets of eyebrows, you can choose one set or the other, but not mix them. So for eyebrows, you have two options: either the first set or the second set. Similarly for other parts?

Wait, but let me check. The problem says "2 sets of eyebrows". So maybe each set is a pair? So each set is a pair, so you can't mix left and right. So for eyebrows, you have two options: choose one of the two sets. Similarly for ears, lips, etc.?

But let me check the other parts. For example, the eyes: "1 pair of googly eyes". So that's a single pair. So you can't choose different eyes? So you have only one option for eyes. Similarly, shoes: regular shoes and clown shoes. So two options for shoes. So shoes are two different pairs? So you can choose either regular or clown shoes. So that's two options.

So, the question is about how many different personalities Tom can make. A complete personality includes eyebrows, eyes, ears, lips, shoes, and optionally hair. So hair is optional. So we need to calculate the number of combinations considering all these parts.

So let me break it down:

First, let's consider each component:

1. Eyebrows: 2 sets. So you have to choose one set. So 2 options.

Wait, but the problem says "you cannot mix and match". So if you have two sets of eyebrows, does that mean you can choose one set or the other? So yes, 2 choices.

2. Eyes: 1 pair. So only 1 option. So you have to include that. So 1 choice.

3. Ears: 2 sets. So similar to eyebrows. So 2 choices.

4. Lips: 2 sets. So 2 choices.

5. Shoes: 1 pair regular, 1 pair clown. So 2 choices.

6. Hair: 3 hairstyles, and it's optional. So for hair, you can choose one of the 3 hairstyles or choose none (bald). So that's 3 + 1 = 4 choices.

Wait, but the problem says "optionally hair". So hair is optional. So the number of choices for hair is 3 (the hairstyles) plus 1 (no hair). So 4 options.

Therefore, the total number of different personalities would be the product of the number of choices for each component.

So that would be:

Eyebrows (2) * Eyes (1) * Ears (2) * Lips (2) * Shoes (2) * Hair (4).

Let me compute that:

First, 2 * 1 = 2.

Then 2 * 2 = 4.

Then 4 * 2 = 8.

Then 8 * 2 = 16.

Then 16 * 4 = 64.

Wait, so 64 different personalities?

But let me check again. Let me make sure I didn't miss anything.

Wait, the problem says "a complete Mr. Potato Head personality includes eyebrows, eyes, ears, lips, shoes and optionally hair". So all the components except hair are required. So yes, eyebrows, eyes, ears, lips, shoes are required. Hair is optional. So that's correct.

So for each of those required parts:

Eyebrows: 2 choices.

Eyes: 1 choice.

Ears: 2 choices.

Lips: 2 choices.

Shoes: 2 choices.

Hair: 4 choices (including none).

So total combinations: 2 * 1 * 2 * 2 * 2 * 4.

Let me compute that again:

2 (eyebrows) * 1 (eyes) = 2.

2 * 2 (ears) = 4.

4 * 2 (lips) = 8.

8 * 2 (shoes) = 16.

16 * 4 (hair) = 64.

So 64 different personalities. Is that correct?

Wait, but let me check if there's something I missed. For example, the note says "you cannot mix and match". So for example, for eyebrows, since there are two sets, each set is a pair. So you can't take one eyebrow from one set and another from another set. So that means that for eyebrows, you have to choose one entire set. So that's 2 options. Similarly for ears and lips. So that's correct.

For eyes, there's only one pair, so you have to take that. So 1 option. For shoes, there are two pairs: regular and clown. So 2 options. For hair, 3 options plus the option of none. So 4. So yes, that seems right.

Wait, but let me think again. Is there a possibility that the hair is considered as a separate part? For example, if you have 3 hairstyles, but maybe each hairstyle is a single piece? But the problem says "3 hairstyles", so I think each is a separate option. So you can choose one of the three or none. So 4 options. So that seems okay.

Alternatively, maybe the problem is considering that the hair is optional, so if you don't choose hair, that's one possibility. So 3 + 1 = 4. So that's correct.

So then, multiplying all together, 2*1*2*2*2*4 = 64. So 64 different personalities.

But let me check again. Maybe I made a mistake in the number of options for each part.

Eyebrows: 2 sets. So 2 choices. Correct.

Eyes: 1 pair. So only 1 choice. Correct.

Ears: 2 sets. So 2 choices. Correct.

Lips: 2 sets. 2 choices. Correct.

Shoes: 1 regular pair, 1 clown pair. So 2 choices. Correct.

Hair: 3 styles, or none. So 4 choices. Correct.

So 2*1*2*2*2*4. Let me compute that again:

First, 2*1 = 2.

Then 2*2 = 4.

Then 4*2 = 8.

Then 8*2 = 16.

Then 16*4 = 64. Yes, same as before.

So 64. So is that the answer? Wait, but let me check again if there's something else.

Wait, the problem says "a complete Mr. Potato Head personality includes eyebrows, eyes, ears, lips, shoes and optionally hair". So all the required parts are included. So the only optional part is hair. So yes, that's accounted for.

But let me think again about the eyes. The problem says "1 pair of googly eyes". So that's a single pair. So you have to include that. So only one option. So that's correct.

Another thing: the problem says "you cannot 'mix and match'". So for example, if there were multiple parts that could be mixed, but in this case, for eyebrows, since they are two sets, each set is a pair. So you can't take one eyebrow from one set and another from another set. So you have to choose one entire set. So that's 2 options. Similarly for ears and lips. So that's correct.

So I think that's all. Therefore, the total number is 64. So the answer is 64. Therefore, \boxed{64}.

Wait, but let me check again. Maybe I made a mistake in the multiplication. Let me do it step by step:

Number of choices:

Eyebrows: 2

Eyes: 1

Ears: 2

Lips: 2

Shoes: 2

Hair: 4

Total combinations: 2 * 1 * 2 * 2 * 2 * 4.

Let me compute 2*1 first: 2.

Then 2*2 (ears): 4.

Then 4*2 (lips): 8.

Then 8*2 (shoes): 16.

Then 16*4 (hair): 64. Yes, same result.

Alternatively, maybe the problem considers that the hair is not optional? Wait, no. The problem says "optionally hair". So hair is optional. So that's correct.

Alternatively, maybe the problem counts the hair as part of the personality? Wait, the problem says "a complete Mr. Potato Head personality includes eyebrows, eyes, ears, lips, shoes and optionally hair". So hair is optional. So if you don't have hair, it's still a complete personality. So that's correct.

Therefore, I think 64 is the correct answer. So I think the answer is 64. Therefore, \boxed{64}.

**Final Answer**
\boxed{64}

To determine how many different **wacky personalities** Tom can create with his Mr. Potato Head, we need to analyze the components he has and how they can be combined, keeping in mind the constraints provided.

---

### **Step 1: Understand the Components and Their Options**

Here are the parts and the number of choices for each:

- **Eyebrows**: 2 sets → 2 choices (choose one set)
- **Eyes**: 1 pair of googly eyes → 1 choice (only one option)
- **Ears**: 2 sets → 2 choices
- **Lips**: 2 sets → 2 choices
- **Shoes**: 1 pair of regular shoes and 1 pair of clown shoes → 2 choices
- **Hair**: 3 hairstyles, and it's **optional** → 3 + 1 = 4 choices (include none)

---

### **Step 2: Determine the Total Number of Combinations**

Since each component is independent and the choices are multiplicative, we multiply the number of choices for each part:

$$
\text{Total combinations} = 2 \times 1 \times 2 \times 2 \times 2 \times 4
$$

Let’s compute this step by step:

- $2 \times 1 = 2$
- $2 \times 2 = 4$
- $4 \times 2 = 8$
- $8 \times 2 = 16$
- $16 \times 4 = 64$

---

### **Step 3: Final Answer**

Thus, the total number of different **wacky personalities** Tom can create is:

$$
\boxed{64}
$$

[Assistant]
Explanation:
Okay, so Tom has this Mr. Potato Head, and he wants to figure out how many different wacky personalities he can make. Let me try to break this down step by step. First, let me list out all the parts he has:

- 3 hairstyles
- 2 sets of eyebrows
- 1 pair of googly eyes
- 2 sets of ears
- 2 sets of lips
- 1 pair of regular shoes
- 1 bonus pair of clown shoes

And the question says a complete personality includes eyebrows, eyes, ears, lips, shoes, and optionally hair. Also, note that you can't mix and match parts, like you can't take one eyebrow from one pair and another from another pair. So each set is considered as a whole, I guess? Or maybe each set has left and right? Wait, the note says you can't mix and match, so maybe each set is a pair? For example, the eyebrows are two sets, so maybe each set is a pair? Wait, but the problem says "2 sets of eyebrows". Hmm. Let me think.

Wait, the problem says "you cannot 'mix and match'. For example, you cannot take the left eyebrow from one pair and the right eyebrow from the other pair." So that suggests that each set of eyebrows is a pair, meaning that each set has a left and right eyebrow. So if you have two sets of eyebrows, you can choose one set or the other, but not mix them. So for eyebrows, you have two options: either the first set or the second set. Similarly for other parts?

Wait, but let me check. The problem says "2 sets of eyebrows". So maybe each set is a pair? So each set is a pair, so you can't mix left and right. So for eyebrows, you have two options: choose one of the two sets. Similarly for ears, lips, etc.?

But let me check the other parts. For example, the eyes: "1 pair of googly eyes". So that's a single pair. So you can't choose different eyes? So you have only one option for eyes. Similarly, shoes: regular shoes and clown shoes. So two options for shoes. So shoes are two different pairs? So you can choose either regular or clown shoes. So that's two options.

So, the question is about how many different personalities Tom can make. A complete personality includes eyebrows, eyes, ears, lips, shoes, and optionally hair. So hair is optional. So we need to calculate the number of combinations considering all these parts.

So let me break it down:

First, let's consider each component:

1. Eyebrows: 2 sets. So you have to choose one set. So 2 options.

Wait, but the problem says "you cannot mix and match". So if you have two sets of eyebrows, does that mean you can choose one set or the other? So yes, 2 choices.

2. Eyes: 1 pair. So only 1 option. So you have to include that. So 1 choice.

3. Ears: 2 sets. So similar to eyebrows. So 2 choices.

4. Lips: 2 sets. So 2 choices.

5. Shoes: 1 pair regular, 1 pair clown. So 2 choices.

6. Hair: 3 hairstyles, and it's optional. So for hair, you can choose one of the 3 hairstyles or choose none (bald). So that's 3 + 1 = 4 choices.

Wait, but the problem says "optionally hair". So hair is optional. So the number of choices for hair is 3 (the hairstyles) plus 1 (no hair). So 4 options.

Therefore, the total number of different personalities would be the product of the number of choices for each component.

So that would be:

Eyebrows (2) * Eyes (1) * Ears (2) * Lips (2) * Shoes (2) * Hair (4).

Let me compute that:

First, 2 * 1 = 2.

Then 2 * 2 = 4.

Then 4 * 2 = 8.

Then 8 * 2 = 16.

Then 16 * 4 = 64.

Wait, so 64 different personalities?

But let me check again. Let me make sure I didn't miss anything.

Wait, the problem says "a complete Mr. Potato Head personality includes eyebrows, eyes, ears, lips, shoes and optionally hair". So all the components except hair are required. So yes, eyebrows, eyes, ears, lips, shoes are required. Hair is optional. So that's correct.

So for each of those required parts:

Eyebrows: 2 choices.

Eyes: 1 choice.

Ears: 2 choices.

Lips: 2 choices.

Shoes: 2 choices.

Hair: 4 choices (including none).

So total combinations: 2 * 1 * 2 * 2 * 2 * 4.

Let me compute that again:

2 (eyebrows) * 1 (eyes) = 2.

2 * 2 (ears) = 4.

4 * 2 (lips) = 8.

8 * 2 (shoes) = 16.

16 * 4 (hair) = 64.

So 64 different personalities. Is that correct?

Wait, but let me check if there's something I missed. For example, the note says "you cannot mix and match". So for example, for eyebrows, since there are two sets, each set is a pair. So you can't take one eyebrow from one set and another from another set. So that means that for eyebrows, you have to choose one entire set. So that's 2 options. Similarly for ears and lips. So that's correct.

For eyes, there's only one pair, so you have to take that. So 1 option. For shoes, there are two pairs: regular and clown. So 2 options. For hair, 3 options plus the option of none. So 4. So yes, that seems right.

Wait, but let me think again. Is there a possibility that the hair is considered as a separate part? For example, if you have 3 hairstyles, but maybe each hairstyle is a single piece? But the problem says "3 hairstyles", so I think each is a separate option. So you can choose one of the three or none. So 4 options. So that seems okay.

Alternatively, maybe the problem is considering that the hair is optional, so if you don't choose hair, that's one possibility. So 3 + 1 = 4. So that's correct.

So then, multiplying all together, 2*1*2*2*2*4 = 64. So 64 different personalities.

But let me check again. Maybe I made a mistake in the number of options for each part.

Eyebrows: 2 sets. So 2 choices. Correct.

Eyes: 1 pair. So only 1 choice. Correct.

Ears: 2 sets. So 2 choices. Correct.

Lips: 2 sets. 2 choices. Correct.

Shoes: 1 regular pair, 1 clown pair. So 2 choices. Correct.

Hair: 3 styles, or none. So 4 choices. Correct.

So 2*1*2*2*2*4. Let me compute that again:

First, 2*1 = 2.

Then 2*2 = 4.

Then 4*2 = 8.

Then 8*2 = 16.

Then 16*4 = 64. Yes, same as before.

So 64. So is that the answer? Wait, but let me check again if there's something else.

Wait, the problem says "a complete Mr. Potato Head personality includes eyebrows, eyes, ears, lips, shoes and optionally hair". So all the required parts are included. So the only optional part is hair. So yes, that's accounted for.

But let me think again about the eyes. The problem says "1 pair of googly eyes". So that's a single pair. So you have to include that. So only one option. So that's correct.

Another thing: the problem says "you cannot 'mix and match'". So for example, if there were multiple parts that could be mixed, but in this case, for eyebrows, since they are two sets, each set is a pair. So you can't take one eyebrow from one set and another from another set. So you have to choose one entire set. So that's 2 options. Similarly for ears and lips. So that's correct.

So I think that's all. Therefore, the total number is 64. So the answer is 64. Therefore, \boxed{64}.

Wait, but let me check again. Maybe I made a mistake in the multiplication. Let me do it step by step:

Number of choices:

Eyebrows: 2

Eyes: 1

Ears: 2

Lips: 2

Shoes: 2

Hair: 4

Total combinations: 2 * 1 * 2 * 2 * 2 * 4.

Let me compute 2*1 first: 2.

Then 2*2 (ears): 4.

Then 4*2 (lips): 8.

Then 8*2 (shoes): 16.

Then 16*4 (hair): 64. Yes, same result.

Alternatively, maybe the problem considers that the hair is not optional? Wait, no. The problem says "optionally hair". So hair is optional. So that's correct.

Alternatively, maybe the problem counts the hair as part of the personality? Wait, the problem says "a complete Mr. Potato Head personality includes eyebrows, eyes, ears, lips, shoes and optionally hair". So hair is optional. So if you don't have hair, it's still a complete personality. So that's correct.

Therefore, I think 64 is the correct answer. So I think the answer is 64. Therefore, \boxed{64}.

**Final Answer**
\boxed{64}

To determine how many different **wacky personalities** Tom can create with his Mr. Potato Head, we need to analyze the components he has and how they can be combined, keeping in mind the constraints provided.

---

### **Step 1: Understand the Components and Their Options**

Here are the parts and the number of choices for each:

- **Eyebrows**: 2 sets → 2 choices (choose one set)
- **Eyes**: 1 pair of googly eyes → 1 choice (only one option)
- **Ears**: 2 sets → 2 choices
- **Lips**: 2 sets → 2 choices
- **Shoes**: 1 pair of regular shoes and 1 pair of clown shoes → 2 choices
- **Hair**: 3 hairstyles, and it's **optional** → 3 + 1 = 4 choices (include none)

---

### **Step 2: Determine the Total Number of Combinations**

Since each component is independent and the choices are multiplicative, we multiply the number of choices for each part:

$$
\text{Total combinations} = 2 \times 1 \times 2 \times 2 \times 2 \times 4
$$

Let’s compute this step by step:

- $2 \times 1 = 2$
- $2 \times 2 = 4$
- $4 \times 2 = 8$
- $8 \times 2 = 16$
- $16 \times 4 = 64$

---

### **Step 3: Final Answer**

Thus, the total number of different **wacky personalities** Tom can create is:

$$
\boxed{64}
$$

Solution: There are 2 choices of eyebrows, 1 choice for eyes, 2 choices for ears, 2 choices for lips, 2 choices for shoes, and 4 choices for hair (3 choices of hairstyles, or bald), and each can be picked independent of the others, so the number of combinations is the product of those, $2\cdot 1\cdot 2\cdot 2\cdot 2\cdot 4 = \boxed{64}$.


Explanation:
Okay, so I need to figure out how to express $(\sqrt{2} + \sqrt{3})^3$ in the form $a\sqrt{2} + b\sqrt{3} + c\sqrt{6}$ where $a$, $b$, and $c$ are integers. Then, I have to find the sum $a + b + c$. Alright, let me start by recalling some algebraic identities or maybe expanding the expression step by step.

First, I know that when you have something like $(a + b)^3$, it can be expanded using the binomial theorem. The formula is $a^3 + 3a^2b + 3ab^2 + b^3$. Maybe I can apply that here. Let me try that.

Let me denote $a = \sqrt{2}$ and $b = \sqrt{3}$. Then, $(\sqrt{2} + \sqrt{3})^3$ would be:

$$
(\sqrt{2})^3 + 3(\sqrt{2})^2(\sqrt{3}) + 3(\sqrt{2})(\sqrt{3})^2 + (\sqrt{3})^3
$$

Okay, let me compute each term one by one.

First term: $(\sqrt{2})^3$. Well, $\sqrt{2}$ cubed is $2^{1/2 \times 3} = 2^{3/2} = 2^{1} \times 2^{1/2} = 2\sqrt{2}$. So that's $2\sqrt{2}$.

Second term: $3(\sqrt{2})^2(\sqrt{3})$. Let me compute $(\sqrt{2})^2$ first. That's 2. Then multiply by $\sqrt{3}$, so 2 times $\sqrt{3}$ is $2\sqrt{3}$. Then multiply by 3: $3 \times 2\sqrt{3} = 6\sqrt{3}$.

Third term: $3(\sqrt{2})(\sqrt{3})^2$. Let me compute $(\sqrt{3})^2$ first, which is 3. Then multiply by $\sqrt{2}$, so that's $3\sqrt{2}$. Then multiply by 3: $3 \times 3\sqrt{2} = 9\sqrt{2}$.

Fourth term: $(\sqrt{3})^3$. Similar to the first term. $\sqrt{3}$ cubed is $3^{3/2} = 3^{1} \times 3^{1/2} = 3\sqrt{3}$. So that's $3\sqrt{3}$.

Now, let me write all the terms together:

First term: $2\sqrt{2}$

Second term: $6\sqrt{3}$

Third term: $9\sqrt{2}$

Fourth term: $3\sqrt{3}$

Now, combine like terms. The terms with $\sqrt{2}$ are the first and third terms: $2\sqrt{2} + 9\sqrt{2} = (2 + 9)\sqrt{2} = 11\sqrt{2}$.

The terms with $\sqrt{3}$ are the second and fourth terms: $6\sqrt{3} + 3\sqrt{3} = (6 + 3)\sqrt{3} = 9\sqrt{3}$.

Wait, but the problem says the expression should be in the form $a\sqrt{2} + b\sqrt{3} + c\sqrt{6}$. But in my calculation, there is no $\sqrt{6}$ term. Did I do something wrong?

Hmm. Let me check my expansion again. Maybe I missed a cross term that involves $\sqrt{6}$?

Wait, hold on. Let me think. When expanding $(\sqrt{2} + \sqrt{3})^3$, maybe I need to consider that when multiplying out, there could be terms that combine to form $\sqrt{6}$. But according to my previous calculation, I only have terms with $\sqrt{2}$ and $\sqrt{3}$. Maybe I made a mistake in the expansion?

Wait, let me try expanding it again step by step. Maybe I should use a different method, like multiplying step by step instead of using the binomial theorem directly. Let me try that.

First, compute $(\sqrt{2} + \sqrt{3})^2$, then multiply by $(\sqrt{2} + \sqrt{3})$ again.

So, first compute $(\sqrt{2} + \sqrt{3})^2$:

$$
(\sqrt{2})^2 + 2 \times \sqrt{2} \times \sqrt{3} + (\sqrt{3})^2 = 2 + 2\sqrt{6} + 3 = 5 + 2\sqrt{6}
$$

Okay, so that's the square. Now, multiply this by $(\sqrt{2} + \sqrt{3})$:

$$
(5 + 2\sqrt{6})(\sqrt{2} + \sqrt{3})
$$

Let me distribute this multiplication:

First, multiply 5 by $\sqrt{2}$: $5\sqrt{2}$

Then, 5 by $\sqrt{3}$: $5\sqrt{3}$

Then, $2\sqrt{6}$ by $\sqrt{2}$: $2\sqrt{6} \times \sqrt{2} = 2\sqrt{12}$

Similarly, $2\sqrt{6}$ by $\sqrt{3}$: $2\sqrt{6} \times \sqrt{3} = 2\sqrt{18}$

Now, simplify those radicals:

$\sqrt{12}$ is $2\sqrt{3}$, so $2\sqrt{12} = 2 \times 2\sqrt{3} = 4\sqrt{3}$

$\sqrt{18}$ is $3\sqrt{2}$, so $2\sqrt{18} = 2 \times 3\sqrt{2} = 6\sqrt{2}$

So putting all the terms together:

$5\sqrt{2} + 5\sqrt{3} + 4\sqrt{3} + 6\sqrt{2}$

Now combine like terms:

For $\sqrt{2}$: $5\sqrt{2} + 6\sqrt{2} = 11\sqrt{2}$

For $\sqrt{3}$: $5\sqrt{3} + 4\sqrt{3} = 9\sqrt{3}$

So again, I get $11\sqrt{2} + 9\sqrt{3}$. Wait, but there's no $\sqrt{6}$ term here. But the problem says it can be written as $a\sqrt{2} + b\sqrt{3} + c\sqrt{6}$. So does that mean that $c = 0$? But maybe I made a mistake here?

Wait, let me check again. Let me verify with another approach. Maybe I can compute $(\sqrt{2} + \sqrt{3})^3$ numerically and see what the value is, then check if it matches with $11\sqrt{2} + 9\sqrt{3}$ or if there's a $\sqrt{6}$ term.

First, compute $\sqrt{2} \approx 1.4142$, $\sqrt{3} \approx 1.732$, so $\sqrt{2} + \sqrt{3} \approx 1.4142 + 1.732 \approx 3.1462$. Then, cubing that: $3.1462^3$. Let me compute that.

First, compute $3.1462^2$. Let me approximate:

$3.1462^2 = (3 + 0.1462)^2 = 9 + 2 \times 3 \times 0.1462 + (0.1462)^2 ≈ 9 + 0.8772 + 0.0214 ≈ 9.8986$. Then multiply by 3.1462:

So 9.8986 * 3.1462 ≈ Let me approximate this. Let me take 9.8986 * 3 = 29.6958, 9.8986 * 0.1462 ≈ Let me compute 9.8986 * 0.1 = 0.98986, 9.8986 * 0.04 = 0.39594, 9.8986 * 0.0062 ≈ approx 0.0613. Adding those: 0.98986 + 0.39594 = 1.3858 + 0.0613 ≈ 1.4471. So total approx 29.6958 + 1.4471 ≈ 31.1429.

Now, let me compute $11\sqrt{2} + 9\sqrt{3}$ numerically. $11 * 1.4142 ≈ 15.5562$, $9 * 1.732 ≈ 15.588$. Adding them together: 15.5562 + 15.588 ≈ 31.1442. Which is very close to the previous value of approx 31.1429. So that seems to check out. So maybe there is no $\sqrt{6}$ term? But the problem says it can be written in the form with $c\sqrt{6}$. So maybe c is zero? But then why does the problem mention that form? Maybe there's a mistake in my calculation?

Wait, let me check again. Let me try expanding $(\sqrt{2} + \sqrt{3})^3$ using another method. Let me use the binomial expansion again, but maybe I made a mistake in the first calculation.

Original expansion:

$$
(\sqrt{2} + \sqrt{3})^3 = (\sqrt{2})^3 + 3(\sqrt{2})^2(\sqrt{3}) + 3(\sqrt{2})(\sqrt{3})^2 + (\sqrt{3})^3
$$

Calculating each term:

First term: $(\sqrt{2})^3 = 2^{3/2} = 2 \times \sqrt{2}$

Second term: $3 \times (\sqrt{2})^2 \times \sqrt{3} = 3 \times 2 \times \sqrt{3} = 6\sqrt{3}$

Third term: $3 \times \sqrt{2} \times (\sqrt{3})^2 = 3 \times \sqrt{2} \times 3 = 9\sqrt{2}$

Fourth term: $(\sqrt{3})^3 = 3^{3/2} = 3 \times \sqrt{3}$

Adding them up:

First term + third term: $2\sqrt{2} + 9\sqrt{2} = 11\sqrt{2}$

Second term + fourth term: $6\sqrt{3} + 3\sqrt{3} = 9\sqrt{3}$

So again, same result. So there is no $\sqrt{6}$ term. Therefore, maybe the problem is correct, and c is zero? But then why does the problem specify the form with $c\sqrt{6}$? Maybe there's a mistake in the problem? Or maybe I need to check again?

Wait, let me think. Maybe there's a different way to expand it where $\sqrt{6}$ comes into play? Let me try multiplying $(\sqrt{2} + \sqrt{3})^3$ as $(\sqrt{2} + \sqrt{3}) \times (\sqrt{2} + \sqrt{3}) \times (\sqrt{2} + \sqrt{3})$ step by step.

First, compute $(\sqrt{2} + \sqrt{3}) \times (\sqrt{2} + \sqrt{3}) = 2 + 2\sqrt{6} + 3 = 5 + 2\sqrt{6}$, as before.

Then, multiply that by $(\sqrt{2} + \sqrt{3})$:

So, $(5 + 2\sqrt{6})(\sqrt{2} + \sqrt{3})$. Let me do this multiplication again:

Multiply 5 by $\sqrt{2}$: 5√2

Multiply 5 by √3: 5√3

Multiply 2√6 by √2: 2√(6*2) = 2√12 = 2*2√3 = 4√3

Multiply 2√6 by √3: 2√(6*3) = 2√18 = 2*3√2 = 6√2

So adding all terms: 5√2 + 5√3 + 4√3 + 6√2 = (5√2 + 6√2) + (5√3 + 4√3) = 11√2 + 9√3. Same result. So again, no √6 term. Therefore, maybe c is zero? But the problem says "can be written in the form a√2 + b√3 + c√6", so maybe c is zero? But then why include that term? Maybe the problem is designed to have c non-zero? Or maybe I made a mistake in the problem statement?

Wait, let me check the problem again. It says: "The number $(\sqrt{2}+\sqrt{3})^3$ can be written in the form $a\sqrt{2} + b\sqrt{3} + c\sqrt{6}$, where $a$, $b$, and $c$ are integers. What is $a+b+c$?"

So according to my calculations, c is zero. Then, a is 11, b is 9, c is 0. Then, a + b + c = 11 + 9 + 0 = 20. But maybe I need to check again?

Alternatively, perhaps there's a different expansion where √6 is present? Wait, but in all my calculations, there is no √6 term. Let me think again. Maybe I need to consider that when expanding $(\sqrt{2} + \sqrt{3})^3$, there is a cross term that combines to √6? Wait, but in the expansion, all terms are either √2, √3, or √6? Wait, no. Wait, when you multiply terms, for example, in the first expansion, when I used binomial theorem, the cross terms are 3*(sqrt(2))^2*sqrt(3) and 3*sqrt(2)*(sqrt(3))^2. But those are 3*2*sqrt(3) and 3*3*sqrt(2). So they are multiples of sqrt(3) and sqrt(2). So there is no sqrt(6) term. So maybe the problem is correct, and c is zero. Then, the answer is 11 + 9 + 0 = 20. But maybe I need to check with another method?

Alternatively, maybe I can use the fact that $(\sqrt{2} + \sqrt{3})^3 + (\sqrt{2} - \sqrt{3})^3$ would have some terms that cancel out? Wait, but that might not be helpful here. Alternatively, maybe I can think of $(\sqrt{2} + \sqrt{3})^3$ as a combination of radicals. Wait, but according to all calculations, there is no √6 term. So maybe the answer is 20?

But let me check with another approach. Let me compute $(\sqrt{2} + \sqrt{3})^3$ using decimal approximations to check if there's a √6 term.

Earlier, I approximated the value as approximately 31.14. Let me compute 11√2 + 9√3:

11 * 1.4142 ≈ 15.5562

9 * 1.732 ≈ 15.588

Adding them gives ≈ 15.5562 + 15.588 ≈ 31.1442, which matches the approximate value of the original expression. Now, if there was a √6 term, say c√6, then let me see what value c would have to be. Suppose the expression is 11√2 + 9√3 + c√6. Let me compute the numerical value of that. Let me compute 11√2 + 9√3 + c√6 ≈ 15.5562 + 15.588 + c * 2.449 (since √6 ≈ 2.449). So total ≈ 31.1442 + 2.449c. But the actual value is approx 31.1429. So if I set 31.1442 + 2.449c ≈ 31.1429, then 2.449c ≈ -0.0013, so c ≈ -0.0013 / 2.449 ≈ -0.0005. Which is practically zero. Therefore, c is zero. Therefore, the answer is 11 + 9 + 0 = 20.

But the problem says "can be written in the form a√2 + b√3 + c√6", so maybe c is allowed to be zero? So then, the answer is 20. But maybe I need to check again if there's a mistake in my calculation?

Alternatively, maybe the problem is expecting me to consider that when expanding, there is a √6 term? Wait, but in all my calculations, there isn't. Wait, let me check once more. Let me use the binomial expansion again:

$$
(\sqrt{2} + \sqrt{3})^3 = \sum_{k=0}^3 \binom{3}{k} (\sqrt{2})^{3 - k} (\sqrt{3})^k
$$

So:

For k=0: $\binom{3}{0} (\sqrt{2})^3 (\sqrt{3})^0 = 1 * 2^{3/2} * 1 = 2\sqrt{2}$

For k=1: $\binom{3}{1} (\sqrt{2})^2 (\sqrt{3})^1 = 3 * 2 * \sqrt{3} = 6\sqrt{3}$

For k=2: $\binom{3}{2} (\sqrt{2})^1 (\sqrt{3})^2 = 3 * \sqrt{2} * 3 = 9\sqrt{2}$

For k=3: $\binom{3}{3} (\sqrt{2})^0 (\sqrt{3})^3 = 1 * 1 * 3^{3/2} = 3\sqrt{3}$

Adding them up: 2√2 + 6√3 + 9√2 + 3√3 = (2 + 9)√2 + (6 + 3)√3 = 11√2 + 9√3. So again, same result. Therefore, there is no √6 term. Therefore, c = 0. Therefore, a + b + c = 11 + 9 + 0 = 20.

But maybe the problem is from a source where they expect a different answer? Wait, let me check with another method. Let me compute $(\sqrt{2} + \sqrt{3})^3$ using the formula for (a + b)^3, but maybe I need to consider that when expanding, there is a cross term?

Wait, no. The expansion is straightforward. Alternatively, maybe I can think of $(\sqrt{2} + \sqrt{3})^3$ as $(\sqrt{2} + \sqrt{3})(\sqrt{2} + \sqrt{3})^2$. Which we already did, and got 11√2 + 9√3. So again, same result.

Alternatively, maybe the problem is written incorrectly? For example, if it was $(\sqrt{2} + \sqrt{3})^2$, then it would be 5 + 2√6, which is in the form a√2 + b√3 + c√6 with a=0, b=0, c=2, but that's not the case here.

Alternatively, maybe the original problem is different? Wait, the user wrote: "The number $(\sqrt{2}+\sqrt{3})^3$ can be written in the form $a\sqrt{2} + b\sqrt{3} + c\sqrt{6}$, where $a$, $b$, and $c$ are integers. What is $a+b+c$?" So that's exactly as given.

Alternatively, maybe I need to check if there's a different interpretation? For example, maybe the question is in a different form? Wait, but the expansion seems correct. So unless there's a miscalculation here, which I can't find, then the answer is 20.

But let me check with another approach. Let me suppose that $(\sqrt{2} + \sqrt{3})^3 = a\sqrt{2} + b\sqrt{3} + c\sqrt{6}$. Let me square both sides? Wait, no, that might complicate. Alternatively, maybe equate the expressions?

Wait, but if I have $(\sqrt{2} + \sqrt{3})^3 = 11\sqrt{2} + 9\sqrt{3}$, then that's already in the form with c=0. Therefore, the answer is 11 + 9 + 0 = 20. Therefore, maybe the answer is 20. But I need to check if there's a different way where c is non-zero. Wait, but how?

Alternatively, maybe the problem is expecting me to consider that when expanding, there is a term with √6? Wait, but in all calculations, there isn't. Wait, unless I made a mistake in simplifying radicals?

Wait, let me check the term $2\sqrt{6} \times \sqrt{2}$. That is 2 * sqrt(6*2) = 2*sqrt(12) = 2*2*sqrt(3) = 4*sqrt(3). Correct. Similarly, $2\sqrt{6} \times \sqrt{3}$ is 2*sqrt(18) = 2*3*sqrt(2) = 6*sqrt(2). Correct. So those terms are accounted for. Therefore, there is no √6 term. Therefore, c is zero.

Alternatively, maybe the problem is a trick question where c is zero? But why would they include the √6 term then? Maybe the problem is from a source where they expect that, but in reality, it's zero. Alternatively, maybe I need to check with another expansion?

Wait, let me think. Suppose I let x = √2 + √3. Then, x^3 = ?

Alternatively, maybe I can find a relation for x. Let me compute x = √2 + √3. Then, x^2 = 5 + 2√6. Then, x^3 = x^2 * x = (5 + 2√6)(√2 + √3). Which is what I did before, leading to 11√2 + 9√3. So again, same result.

Alternatively, maybe I can write 11√2 + 9√3 as a combination with √6? For example, suppose that 11√2 + 9√3 = a√2 + b√3 + c√6. Then, equating coefficients:

For √2: 11 = a

For √3: 9 = b

For √6: 0 = c

Therefore, c = 0. Therefore, the answer is 11 + 9 + 0 = 20. Therefore, unless there's a mistake in my reasoning, which I can't find, the answer is 20.

But maybe I need to check with another method. Let me compute $(\sqrt{2} + \sqrt{3})^3$ using the binomial theorem again, but this time, writing all terms explicitly:

First, $(\sqrt{2} + \sqrt{3})^3 = (\sqrt{2})^3 + 3(\sqrt{2})^2(\sqrt{3}) + 3(\sqrt{2})(\sqrt{3})^2 + (\sqrt{3})^3$

As before:

First term: $2\sqrt{2}$

Second term: $3 * 2 * \sqrt{3} = 6\sqrt{3}$

Third term: $3 * \sqrt{2} * 3 = 9\sqrt{2}$

Fourth term: $3\sqrt{3}$

Adding them up:

$2\sqrt{2} + 6\sqrt{3} + 9\sqrt{2} + 3\sqrt{3} = (2 + 9)\sqrt{2} + (6 + 3)\sqrt{3} = 11\sqrt{2} + 9\sqrt{3}$. So again, same result. Therefore, I think my answer is correct. Therefore, the value of a + b + c is 11 + 9 + 0 = 20.

But just to be thorough, let me check if there's a possibility that the problem meant $(\sqrt{2} + \sqrt{3})^3$ in some other form? For example, if there was a typo and it was supposed to be $(\sqrt{2} + \sqrt{3})^4$ or something else? But the user wrote $(\sqrt{2}+\sqrt{3})^3$. Alternatively, maybe the problem is from a source where they consider that √6 is present? But according to all calculations, it's not.

Alternatively, maybe I need to rationalize or something else? Wait, but the expression is already simplified. So I think my answer is correct. Therefore, I think the answer is 20.

But just to be 100% sure, let me check with exact values. Let me compute $(\sqrt{2} + \sqrt{3})^3$ and see if it's equal to 11√2 + 9√3.

Let me compute $(\sqrt{2} + \sqrt{3})^3$:

First, expand it as $(\sqrt{2} + \sqrt{3}) \times (\sqrt{2} + \sqrt{3}) \times (\sqrt{2} + \sqrt{3})$.

First, compute $(\sqrt{2} + \sqrt{3})^2 = 2 + 2\sqrt{6} + 3 = 5 + 2\sqrt{6}$.

Then, multiply by $(\sqrt{2} + \sqrt{3})$:

(5 + 2√6)(√2 + √3) = 5√2 + 5√3 + 2√6√2 + 2√6√3.

Compute each term:

5√2 + 5√3 + 2√(6*2) + 2√(6*3) = 5√2 + 5√3 + 2√12 + 2√18.

Simplify radicals:

√12 = 2√3, so 2√12 = 4√3.

√18 = 3√2, so 2√18 = 6√2.

Therefore, total expression is 5√2 + 5√3 + 4√3 + 6√2 = (5√2 + 6√2) + (5√3 + 4√3) = 11√2 + 9√3. So again, same result. Therefore, there is no √6 term. Therefore, c = 0. Therefore, the answer is 11 + 9 + 0 = 20.

Therefore, I think the answer is 20. Maybe the problem includes the √6 term just to have three terms, but in reality, c is zero. Therefore, the answer is 20.

**Final Answer**
\boxed{20}

To solve the problem, we are asked to express the number $(\sqrt{2} + \sqrt{3})^3$ in the form:

$$
a\sqrt{2} + b\sqrt{3} + c\sqrt{6}
$$

where $a$, $b$, and $c$ are integers, and then find the sum $a + b + c$.

---

### Step 1: Expand the Expression

We use the **binomial theorem** to expand $(\sqrt{2} + \sqrt{3})^3$:

$$
(\sqrt{2} + \sqrt{3})^3 = (\sqrt{2})^3 + 3(\sqrt{2})^2(\sqrt{3}) + 3(\sqrt{2})(\sqrt{3})^2 + (\sqrt{3})^3
$$

Now compute each term:

- $(\sqrt{2})^3 = 2\sqrt{2}$
- $3(\sqrt{2})^2(\sqrt{3}) = 3 \cdot 2 \cdot \sqrt{3} = 6\sqrt{3}$
- $3(\sqrt{2})(\sqrt{3})^2 = 3 \cdot \sqrt{2} \cdot 3 = 9\sqrt{2}$
- $(\sqrt{3})^3 = 3\sqrt{3}$

Now combine like terms:

- Terms with $\sqrt{2}$: $2\sqrt{2} + 9\sqrt{2} = 11\sqrt{2}$
- Terms with $\sqrt{3}$: $6\sqrt{3} + 3\sqrt{3} = 9\sqrt{3}$

So the expression becomes:

$$
11\sqrt{2} + 9\sqrt{3}
$$

---

### Step 2: Match the Form

We are told the expression can be written as:

$$
a\sqrt{2} + b\sqrt{3} + c\sqrt{6}
$$

From our expansion, we have:

- $a = 11$
- $b = 9$
- $c = 0$ (since there is no $\sqrt{6}$ term)

---

### Step 3: Compute the Sum

$$
a + b + c = 11 + 9 + 0 = \boxed{20}
$$

---

### Final Answer

$$
\boxed{20}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how to express $(\sqrt{2} + \sqrt{3})^3$ in the form $a\sqrt{2} + b\sqrt{3} + c\sqrt{6}$ where $a$, $b$, and $c$ are integers. Then, I have to find the sum $a + b + c$. Alright, let me start by recalling some algebraic identities or maybe expanding the expression step by step.

First, I know that when you have something like $(a + b)^3$, it can be expanded using the binomial theorem. The formula is $a^3 + 3a^2b + 3ab^2 + b^3$. Maybe I can apply that here. Let me try that.

Let me denote $a = \sqrt{2}$ and $b = \sqrt{3}$. Then, $(\sqrt{2} + \sqrt{3})^3$ would be:

$$
(\sqrt{2})^3 + 3(\sqrt{2})^2(\sqrt{3}) + 3(\sqrt{2})(\sqrt{3})^2 + (\sqrt{3})^3
$$

Okay, let me compute each term one by one.

First term: $(\sqrt{2})^3$. Well, $\sqrt{2}$ cubed is $2^{1/2 \times 3} = 2^{3/2} = 2^{1} \times 2^{1/2} = 2\sqrt{2}$. So that's $2\sqrt{2}$.

Second term: $3(\sqrt{2})^2(\sqrt{3})$. Let me compute $(\sqrt{2})^2$ first. That's 2. Then multiply by $\sqrt{3}$, so 2 times $\sqrt{3}$ is $2\sqrt{3}$. Then multiply by 3: $3 \times 2\sqrt{3} = 6\sqrt{3}$.

Third term: $3(\sqrt{2})(\sqrt{3})^2$. Let me compute $(\sqrt{3})^2$ first, which is 3. Then multiply by $\sqrt{2}$, so that's $3\sqrt{2}$. Then multiply by 3: $3 \times 3\sqrt{2} = 9\sqrt{2}$.

Fourth term: $(\sqrt{3})^3$. Similar to the first term. $\sqrt{3}$ cubed is $3^{3/2} = 3^{1} \times 3^{1/2} = 3\sqrt{3}$. So that's $3\sqrt{3}$.

Now, let me write all the terms together:

First term: $2\sqrt{2}$

Second term: $6\sqrt{3}$

Third term: $9\sqrt{2}$

Fourth term: $3\sqrt{3}$

Now, combine like terms. The terms with $\sqrt{2}$ are the first and third terms: $2\sqrt{2} + 9\sqrt{2} = (2 + 9)\sqrt{2} = 11\sqrt{2}$.

The terms with $\sqrt{3}$ are the second and fourth terms: $6\sqrt{3} + 3\sqrt{3} = (6 + 3)\sqrt{3} = 9\sqrt{3}$.

Wait, but the problem says the expression should be in the form $a\sqrt{2} + b\sqrt{3} + c\sqrt{6}$. But in my calculation, there is no $\sqrt{6}$ term. Did I do something wrong?

Hmm. Let me check my expansion again. Maybe I missed a cross term that involves $\sqrt{6}$?

Wait, hold on. Let me think. When expanding $(\sqrt{2} + \sqrt{3})^3$, maybe I need to consider that when multiplying out, there could be terms that combine to form $\sqrt{6}$. But according to my previous calculation, I only have terms with $\sqrt{2}$ and $\sqrt{3}$. Maybe I made a mistake in the expansion?

Wait, let me try expanding it again step by step. Maybe I should use a different method, like multiplying step by step instead of using the binomial theorem directly. Let me try that.

First, compute $(\sqrt{2} + \sqrt{3})^2$, then multiply by $(\sqrt{2} + \sqrt{3})$ again.

So, first compute $(\sqrt{2} + \sqrt{3})^2$:

$$
(\sqrt{2})^2 + 2 \times \sqrt{2} \times \sqrt{3} + (\sqrt{3})^2 = 2 + 2\sqrt{6} + 3 = 5 + 2\sqrt{6}
$$

Okay, so that's the square. Now, multiply this by $(\sqrt{2} + \sqrt{3})$:

$$
(5 + 2\sqrt{6})(\sqrt{2} + \sqrt{3})
$$

Let me distribute this multiplication:

First, multiply 5 by $\sqrt{2}$: $5\sqrt{2}$

Then, 5 by $\sqrt{3}$: $5\sqrt{3}$

Then, $2\sqrt{6}$ by $\sqrt{2}$: $2\sqrt{6} \times \sqrt{2} = 2\sqrt{12}$

Similarly, $2\sqrt{6}$ by $\sqrt{3}$: $2\sqrt{6} \times \sqrt{3} = 2\sqrt{18}$

Now, simplify those radicals:

$\sqrt{12}$ is $2\sqrt{3}$, so $2\sqrt{12} = 2 \times 2\sqrt{3} = 4\sqrt{3}$

$\sqrt{18}$ is $3\sqrt{2}$, so $2\sqrt{18} = 2 \times 3\sqrt{2} = 6\sqrt{2}$

So putting all the terms together:

$5\sqrt{2} + 5\sqrt{3} + 4\sqrt{3} + 6\sqrt{2}$

Now combine like terms:

For $\sqrt{2}$: $5\sqrt{2} + 6\sqrt{2} = 11\sqrt{2}$

For $\sqrt{3}$: $5\sqrt{3} + 4\sqrt{3} = 9\sqrt{3}$

So again, I get $11\sqrt{2} + 9\sqrt{3}$. Wait, but there's no $\sqrt{6}$ term here. But the problem says it can be written as $a\sqrt{2} + b\sqrt{3} + c\sqrt{6}$. So does that mean that $c = 0$? But maybe I made a mistake here?

Wait, let me check again. Let me verify with another approach. Maybe I can compute $(\sqrt{2} + \sqrt{3})^3$ numerically and see what the value is, then check if it matches with $11\sqrt{2} + 9\sqrt{3}$ or if there's a $\sqrt{6}$ term.

First, compute $\sqrt{2} \approx 1.4142$, $\sqrt{3} \approx 1.732$, so $\sqrt{2} + \sqrt{3} \approx 1.4142 + 1.732 \approx 3.1462$. Then, cubing that: $3.1462^3$. Let me compute that.

First, compute $3.1462^2$. Let me approximate:

$3.1462^2 = (3 + 0.1462)^2 = 9 + 2 \times 3 \times 0.1462 + (0.1462)^2 ≈ 9 + 0.8772 + 0.0214 ≈ 9.8986$. Then multiply by 3.1462:

So 9.8986 * 3.1462 ≈ Let me approximate this. Let me take 9.8986 * 3 = 29.6958, 9.8986 * 0.1462 ≈ Let me compute 9.8986 * 0.1 = 0.98986, 9.8986 * 0.04 = 0.39594, 9.8986 * 0.0062 ≈ approx 0.0613. Adding those: 0.98986 + 0.39594 = 1.3858 + 0.0613 ≈ 1.4471. So total approx 29.6958 + 1.4471 ≈ 31.1429.

Now, let me compute $11\sqrt{2} + 9\sqrt{3}$ numerically. $11 * 1.4142 ≈ 15.5562$, $9 * 1.732 ≈ 15.588$. Adding them together: 15.5562 + 15.588 ≈ 31.1442. Which is very close to the previous value of approx 31.1429. So that seems to check out. So maybe there is no $\sqrt{6}$ term? But the problem says it can be written in the form with $c\sqrt{6}$. So maybe c is zero? But then why does the problem mention that form? Maybe there's a mistake in my calculation?

Wait, let me check again. Let me try expanding $(\sqrt{2} + \sqrt{3})^3$ using another method. Let me use the binomial expansion again, but maybe I made a mistake in the first calculation.

Original expansion:

$$
(\sqrt{2} + \sqrt{3})^3 = (\sqrt{2})^3 + 3(\sqrt{2})^2(\sqrt{3}) + 3(\sqrt{2})(\sqrt{3})^2 + (\sqrt{3})^3
$$

Calculating each term:

First term: $(\sqrt{2})^3 = 2^{3/2} = 2 \times \sqrt{2}$

Second term: $3 \times (\sqrt{2})^2 \times \sqrt{3} = 3 \times 2 \times \sqrt{3} = 6\sqrt{3}$

Third term: $3 \times \sqrt{2} \times (\sqrt{3})^2 = 3 \times \sqrt{2} \times 3 = 9\sqrt{2}$

Fourth term: $(\sqrt{3})^3 = 3^{3/2} = 3 \times \sqrt{3}$

Adding them up:

First term + third term: $2\sqrt{2} + 9\sqrt{2} = 11\sqrt{2}$

Second term + fourth term: $6\sqrt{3} + 3\sqrt{3} = 9\sqrt{3}$

So again, same result. So there is no $\sqrt{6}$ term. Therefore, maybe the problem is correct, and c is zero? But then why does the problem specify the form with $c\sqrt{6}$? Maybe there's a mistake in the problem? Or maybe I need to check again?

Wait, let me think. Maybe there's a different way to expand it where $\sqrt{6}$ comes into play? Let me try multiplying $(\sqrt{2} + \sqrt{3})^3$ as $(\sqrt{2} + \sqrt{3}) \times (\sqrt{2} + \sqrt{3}) \times (\sqrt{2} + \sqrt{3})$ step by step.

First, compute $(\sqrt{2} + \sqrt{3}) \times (\sqrt{2} + \sqrt{3}) = 2 + 2\sqrt{6} + 3 = 5 + 2\sqrt{6}$, as before.

Then, multiply that by $(\sqrt{2} + \sqrt{3})$:

So, $(5 + 2\sqrt{6})(\sqrt{2} + \sqrt{3})$. Let me do this multiplication again:

Multiply 5 by $\sqrt{2}$: 5√2

Multiply 5 by √3: 5√3

Multiply 2√6 by √2: 2√(6*2) = 2√12 = 2*2√3 = 4√3

Multiply 2√6 by √3: 2√(6*3) = 2√18 = 2*3√2 = 6√2

So adding all terms: 5√2 + 5√3 + 4√3 + 6√2 = (5√2 + 6√2) + (5√3 + 4√3) = 11√2 + 9√3. Same result. So again, no √6 term. Therefore, maybe c is zero? But the problem says "can be written in the form a√2 + b√3 + c√6", so maybe c is zero? But then why include that term? Maybe the problem is designed to have c non-zero? Or maybe I made a mistake in the problem statement?

Wait, let me check the problem again. It says: "The number $(\sqrt{2}+\sqrt{3})^3$ can be written in the form $a\sqrt{2} + b\sqrt{3} + c\sqrt{6}$, where $a$, $b$, and $c$ are integers. What is $a+b+c$?"

So according to my calculations, c is zero. Then, a is 11, b is 9, c is 0. Then, a + b + c = 11 + 9 + 0 = 20. But maybe I need to check again?

Alternatively, perhaps there's a different expansion where √6 is present? Wait, but in all my calculations, there is no √6 term. Let me think again. Maybe I need to consider that when expanding $(\sqrt{2} + \sqrt{3})^3$, there is a cross term that combines to √6? Wait, but in the expansion, all terms are either √2, √3, or √6? Wait, no. Wait, when you multiply terms, for example, in the first expansion, when I used binomial theorem, the cross terms are 3*(sqrt(2))^2*sqrt(3) and 3*sqrt(2)*(sqrt(3))^2. But those are 3*2*sqrt(3) and 3*3*sqrt(2). So they are multiples of sqrt(3) and sqrt(2). So there is no sqrt(6) term. So maybe the problem is correct, and c is zero. Then, the answer is 11 + 9 + 0 = 20. But maybe I need to check with another method?

Alternatively, maybe I can use the fact that $(\sqrt{2} + \sqrt{3})^3 + (\sqrt{2} - \sqrt{3})^3$ would have some terms that cancel out? Wait, but that might not be helpful here. Alternatively, maybe I can think of $(\sqrt{2} + \sqrt{3})^3$ as a combination of radicals. Wait, but according to all calculations, there is no √6 term. So maybe the answer is 20?

But let me check with another approach. Let me compute $(\sqrt{2} + \sqrt{3})^3$ using decimal approximations to check if there's a √6 term.

Earlier, I approximated the value as approximately 31.14. Let me compute 11√2 + 9√3:

11 * 1.4142 ≈ 15.5562

9 * 1.732 ≈ 15.588

Adding them gives ≈ 15.5562 + 15.588 ≈ 31.1442, which matches the approximate value of the original expression. Now, if there was a √6 term, say c√6, then let me see what value c would have to be. Suppose the expression is 11√2 + 9√3 + c√6. Let me compute the numerical value of that. Let me compute 11√2 + 9√3 + c√6 ≈ 15.5562 + 15.588 + c * 2.449 (since √6 ≈ 2.449). So total ≈ 31.1442 + 2.449c. But the actual value is approx 31.1429. So if I set 31.1442 + 2.449c ≈ 31.1429, then 2.449c ≈ -0.0013, so c ≈ -0.0013 / 2.449 ≈ -0.0005. Which is practically zero. Therefore, c is zero. Therefore, the answer is 11 + 9 + 0 = 20.

But the problem says "can be written in the form a√2 + b√3 + c√6", so maybe c is allowed to be zero? So then, the answer is 20. But maybe I need to check again if there's a mistake in my calculation?

Alternatively, maybe the problem is expecting me to consider that when expanding, there is a √6 term? Wait, but in all my calculations, there isn't. Wait, let me check once more. Let me use the binomial expansion again:

$$
(\sqrt{2} + \sqrt{3})^3 = \sum_{k=0}^3 \binom{3}{k} (\sqrt{2})^{3 - k} (\sqrt{3})^k
$$

So:

For k=0: $\binom{3}{0} (\sqrt{2})^3 (\sqrt{3})^0 = 1 * 2^{3/2} * 1 = 2\sqrt{2}$

For k=1: $\binom{3}{1} (\sqrt{2})^2 (\sqrt{3})^1 = 3 * 2 * \sqrt{3} = 6\sqrt{3}$

For k=2: $\binom{3}{2} (\sqrt{2})^1 (\sqrt{3})^2 = 3 * \sqrt{2} * 3 = 9\sqrt{2}$

For k=3: $\binom{3}{3} (\sqrt{2})^0 (\sqrt{3})^3 = 1 * 1 * 3^{3/2} = 3\sqrt{3}$

Adding them up: 2√2 + 6√3 + 9√2 + 3√3 = (2 + 9)√2 + (6 + 3)√3 = 11√2 + 9√3. So again, same result. Therefore, there is no √6 term. Therefore, c = 0. Therefore, a + b + c = 11 + 9 + 0 = 20.

But maybe the problem is from a source where they expect a different answer? Wait, let me check with another method. Let me compute $(\sqrt{2} + \sqrt{3})^3$ using the formula for (a + b)^3, but maybe I need to consider that when expanding, there is a cross term?

Wait, no. The expansion is straightforward. Alternatively, maybe I can think of $(\sqrt{2} + \sqrt{3})^3$ as $(\sqrt{2} + \sqrt{3})(\sqrt{2} + \sqrt{3})^2$. Which we already did, and got 11√2 + 9√3. So again, same result.

Alternatively, maybe the problem is written incorrectly? For example, if it was $(\sqrt{2} + \sqrt{3})^2$, then it would be 5 + 2√6, which is in the form a√2 + b√3 + c√6 with a=0, b=0, c=2, but that's not the case here.

Alternatively, maybe the original problem is different? Wait, the user wrote: "The number $(\sqrt{2}+\sqrt{3})^3$ can be written in the form $a\sqrt{2} + b\sqrt{3} + c\sqrt{6}$, where $a$, $b$, and $c$ are integers. What is $a+b+c$?" So that's exactly as given.

Alternatively, maybe I need to check if there's a different interpretation? For example, maybe the question is in a different form? Wait, but the expansion seems correct. So unless there's a miscalculation here, which I can't find, then the answer is 20.

But let me check with another approach. Let me suppose that $(\sqrt{2} + \sqrt{3})^3 = a\sqrt{2} + b\sqrt{3} + c\sqrt{6}$. Let me square both sides? Wait, no, that might complicate. Alternatively, maybe equate the expressions?

Wait, but if I have $(\sqrt{2} + \sqrt{3})^3 = 11\sqrt{2} + 9\sqrt{3}$, then that's already in the form with c=0. Therefore, the answer is 11 + 9 + 0 = 20. Therefore, maybe the answer is 20. But I need to check if there's a different way where c is non-zero. Wait, but how?

Alternatively, maybe the problem is expecting me to consider that when expanding, there is a term with √6? Wait, but in all calculations, there isn't. Wait, unless I made a mistake in simplifying radicals?

Wait, let me check the term $2\sqrt{6} \times \sqrt{2}$. That is 2 * sqrt(6*2) = 2*sqrt(12) = 2*2*sqrt(3) = 4*sqrt(3). Correct. Similarly, $2\sqrt{6} \times \sqrt{3}$ is 2*sqrt(18) = 2*3*sqrt(2) = 6*sqrt(2). Correct. So those terms are accounted for. Therefore, there is no √6 term. Therefore, c is zero.

Alternatively, maybe the problem is a trick question where c is zero? But why would they include the √6 term then? Maybe the problem is from a source where they expect that, but in reality, it's zero. Alternatively, maybe I need to check with another expansion?

Wait, let me think. Suppose I let x = √2 + √3. Then, x^3 = ?

Alternatively, maybe I can find a relation for x. Let me compute x = √2 + √3. Then, x^2 = 5 + 2√6. Then, x^3 = x^2 * x = (5 + 2√6)(√2 + √3). Which is what I did before, leading to 11√2 + 9√3. So again, same result.

Alternatively, maybe I can write 11√2 + 9√3 as a combination with √6? For example, suppose that 11√2 + 9√3 = a√2 + b√3 + c√6. Then, equating coefficients:

For √2: 11 = a

For √3: 9 = b

For √6: 0 = c

Therefore, c = 0. Therefore, the answer is 11 + 9 + 0 = 20. Therefore, unless there's a mistake in my reasoning, which I can't find, the answer is 20.

But maybe I need to check with another method. Let me compute $(\sqrt{2} + \sqrt{3})^3$ using the binomial theorem again, but this time, writing all terms explicitly:

First, $(\sqrt{2} + \sqrt{3})^3 = (\sqrt{2})^3 + 3(\sqrt{2})^2(\sqrt{3}) + 3(\sqrt{2})(\sqrt{3})^2 + (\sqrt{3})^3$

As before:

First term: $2\sqrt{2}$

Second term: $3 * 2 * \sqrt{3} = 6\sqrt{3}$

Third term: $3 * \sqrt{2} * 3 = 9\sqrt{2}$

Fourth term: $3\sqrt{3}$

Adding them up:

$2\sqrt{2} + 6\sqrt{3} + 9\sqrt{2} + 3\sqrt{3} = (2 + 9)\sqrt{2} + (6 + 3)\sqrt{3} = 11\sqrt{2} + 9\sqrt{3}$. So again, same result. Therefore, I think my answer is correct. Therefore, the value of a + b + c is 11 + 9 + 0 = 20.

But just to be thorough, let me check if there's a possibility that the problem meant $(\sqrt{2} + \sqrt{3})^3$ in some other form? For example, if there was a typo and it was supposed to be $(\sqrt{2} + \sqrt{3})^4$ or something else? But the user wrote $(\sqrt{2}+\sqrt{3})^3$. Alternatively, maybe the problem is from a source where they consider that √6 is present? But according to all calculations, it's not.

Alternatively, maybe I need to rationalize or something else? Wait, but the expression is already simplified. So I think my answer is correct. Therefore, I think the answer is 20.

But just to be 100% sure, let me check with exact values. Let me compute $(\sqrt{2} + \sqrt{3})^3$ and see if it's equal to 11√2 + 9√3.

Let me compute $(\sqrt{2} + \sqrt{3})^3$:

First, expand it as $(\sqrt{2} + \sqrt{3}) \times (\sqrt{2} + \sqrt{3}) \times (\sqrt{2} + \sqrt{3})$.

First, compute $(\sqrt{2} + \sqrt{3})^2 = 2 + 2\sqrt{6} + 3 = 5 + 2\sqrt{6}$.

Then, multiply by $(\sqrt{2} + \sqrt{3})$:

(5 + 2√6)(√2 + √3) = 5√2 + 5√3 + 2√6√2 + 2√6√3.

Compute each term:

5√2 + 5√3 + 2√(6*2) + 2√(6*3) = 5√2 + 5√3 + 2√12 + 2√18.

Simplify radicals:

√12 = 2√3, so 2√12 = 4√3.

√18 = 3√2, so 2√18 = 6√2.

Therefore, total expression is 5√2 + 5√3 + 4√3 + 6√2 = (5√2 + 6√2) + (5√3 + 4√3) = 11√2 + 9√3. So again, same result. Therefore, there is no √6 term. Therefore, c = 0. Therefore, the answer is 11 + 9 + 0 = 20.

Therefore, I think the answer is 20. Maybe the problem includes the √6 term just to have three terms, but in reality, c is zero. Therefore, the answer is 20.

**Final Answer**
\boxed{20}

To solve the problem, we are asked to express the number $(\sqrt{2} + \sqrt{3})^3$ in the form:

$$
a\sqrt{2} + b\sqrt{3} + c\sqrt{6}
$$

where $a$, $b$, and $c$ are integers, and then find the sum $a + b + c$.

---

### Step 1: Expand the Expression

We use the **binomial theorem** to expand $(\sqrt{2} + \sqrt{3})^3$:

$$
(\sqrt{2} + \sqrt{3})^3 = (\sqrt{2})^3 + 3(\sqrt{2})^2(\sqrt{3}) + 3(\sqrt{2})(\sqrt{3})^2 + (\sqrt{3})^3
$$

Now compute each term:

- $(\sqrt{2})^3 = 2\sqrt{2}$
- $3(\sqrt{2})^2(\sqrt{3}) = 3 \cdot 2 \cdot \sqrt{3} = 6\sqrt{3}$
- $3(\sqrt{2})(\sqrt{3})^2 = 3 \cdot \sqrt{2} \cdot 3 = 9\sqrt{2}$
- $(\sqrt{3})^3 = 3\sqrt{3}$

Now combine like terms:

- Terms with $\sqrt{2}$: $2\sqrt{2} + 9\sqrt{2} = 11\sqrt{2}$
- Terms with $\sqrt{3}$: $6\sqrt{3} + 3\sqrt{3} = 9\sqrt{3}$

So the expression becomes:

$$
11\sqrt{2} + 9\sqrt{3}
$$

---

### Step 2: Match the Form

We are told the expression can be written as:

$$
a\sqrt{2} + b\sqrt{3} + c\sqrt{6}
$$

From our expansion, we have:

- $a = 11$
- $b = 9$
- $c = 0$ (since there is no $\sqrt{6}$ term)

---

### Step 3: Compute the Sum

$$
a + b + c = 11 + 9 + 0 = \boxed{20}
$$

---

### Final Answer

$$
\boxed{20}
$$

Solution:
First, we calculate $(\sqrt{2} + \sqrt{3})^2$: \begin{align*}
(\sqrt{2} + \sqrt{3})^2 &= (\sqrt{2} + \sqrt{3})(\sqrt{2}+\sqrt{3})\\
&=(\sqrt{2})(\sqrt{2}) + (\sqrt{2})(\sqrt{3}) + (\sqrt{3})(\sqrt{2}) + (\sqrt{3})(\sqrt{3})\\
&= 2 + \sqrt{6} + \sqrt{6} + 3\\
&=5+2\sqrt{6}.
\end{align*} Multiplying this by $\sqrt{2} +\sqrt{3}$ gives \begin{align*}
(\sqrt{2}+ \sqrt{3})^3 &=(\sqrt{2}+\sqrt{3})^2 (\sqrt{2} +\sqrt{3})\\
&=(5+2\sqrt{6})(\sqrt{2} +\sqrt{3})\\
&= 5\sqrt{2} + 5\sqrt{3} + (2\sqrt{6})(\sqrt{2}) + (2\sqrt{6})(\sqrt{3})\\
&=5\sqrt{2} + 5\sqrt{3} + 2\sqrt{12} + 2\sqrt{18}\\
&=5\sqrt{2} + 5\sqrt{3} + 2(2\sqrt{3}) + 2(3\sqrt{2})\\
&=11\sqrt{2} + 9\sqrt{3}.
\end{align*} Therefore, we have $a+b+c = \boxed{20}$.  (Notice that $c=0;$ tricky!)

We may also expand $(\sqrt{3} + \sqrt{2})^3$ using the Binomial Theorem, giving us ${\sqrt{2}}^3 + 3{\sqrt{2}}^2\sqrt{3}+3\sqrt{2}{\sqrt{3}}^2+{\sqrt{3}}^3$. Simplifying this yields $2\sqrt{2}+6\sqrt{3}+9\sqrt{2}+3\sqrt{3} = 11\sqrt{2}+9\sqrt{3}$, and once again $a + b + c = \boxed{20}$.