Okay, let's see. I need to find the value of a + b + c where the expression (√2 + √3)^3 is written in the form a√2 + b√3 + c√6. Alright, so first, maybe I should expand the cube of (√2 + √3). Let me recall how to expand a binomial raised to the third power. The formula is (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3. So, substituting a = √2 and b = √3, that should work.

Let me write that out step by step. So, (√2 + √3)^3 = (√2)^3 + 3*(√2)^2*(√3) + 3*(√2)*(√3)^2 + (√3)^3. Now, let's compute each term individually.

First term: (√2)^3. That's (√2)^2 * √2 = 2 * √2 = 2√2. Okay, so that's 2√2.

Second term: 3*(√2)^2*(√3). (√2)^2 is 2, so this becomes 3*2*√3 = 6√3.

Third term: 3*(√2)*(√3)^2. (√3)^2 is 3, so this is 3*√2*3 = 9√2.

Fourth term: (√3)^3. That's (√3)^2 * √3 = 3*√3 = 3√3.

Now, let's add all these terms together. So:

First term: 2√2

Second term: 6√3

Third term: 9√2

Fourth term: 3√3

Adding the like terms. Let's combine the √2 terms: 2√2 + 9√2 = 11√2.

Then the √3 terms: 6√3 + 3√3 = 9√3.

And there's also a √6 term? Wait, hold on. Wait, the original expansion was (√2 + √3)^3. Let me check if there's a cross term that results in √6. Because when you expand (√2 + √3)^3, using the binomial theorem, each term is of the form C(n, k) * (√2)^(n-k) * (√3)^k. Let me see for n=3 and k=1, 2, or 3.

Wait, actually, maybe I made a mistake in the expansion. Let me check again. Wait, when expanding (a + b)^3, the coefficients are 1, 3, 3, 1. So, the expansion is a^3 + 3a^2b + 3ab^2 + b^3. So substituting a = √2 and b = √3:

First term: (√2)^3 = 2√2

Second term: 3*(√2)^2*(√3) = 3*2*√3 = 6√3

Third term: 3*(√2)*(√3)^2 = 3*√2*3 = 9√2

Fourth term: (√3)^3 = 3√3

So, when we add all terms:

2√2 + 6√3 + 9√2 + 3√3

Combine like terms:

(2√2 + 9√2) + (6√3 + 3√3) = 11√2 + 9√3

Wait, but the problem states that the expression can be written in the form a√2 + b√3 + c√6. But in my calculation, there are only √2 and √3 terms, no √6 term. That seems odd. Did I miss something?

Wait, maybe I need to check if there's a cross term when expanding (√2 + √3)^3. Let me try expanding it another way, perhaps using the distributive property step by step.

First, compute (√2 + √3)^2 first, then multiply by (√2 + √3).

Compute (√2 + √3)^2:

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

Then multiply this by (√2 + √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.

Simplify the radicals:

√6*√2 = √(6*2) = √12 = 2√3

√6*√3 = √(6*3) = √18 = 3√2

So substituting back:

5√2 + 5√3 + 2*(2√3) + 2*(3√2) = 5√2 + 5√3 + 4√3 + 6√2.

Combine like terms:

(5√2 + 6√2) + (5√3 + 4√3) = 11√2 + 9√3.

Again, same result. So there's no √6 term. But the problem states that it can be written as a√2 + b√3 + c√6. That suggests that perhaps the original problem might have a typo, or maybe I made a mistake. Wait, let me check the problem again.

Wait, the problem says: "The number (√2 + √3)^3 can be written in the form a√2 + b√3 + c√6, where a, b, and c are integers. What is a + b + c?"

But according to my expansion, there's no √6 term. So either I made a mistake in calculation, or the problem is incorrect. Wait, let me check again.

Wait, when I expanded (√2 + √3)^3, I first expanded it as (√2 + √3)*(√2 + √3)*(√2 + √3). Alternatively, maybe I should use the binomial theorem step by step.

Wait, binomial theorem for (a + b)^3 is a^3 + 3a^2b + 3ab^2 + b^3. Let's apply that with a = √2 and b = √3.

So:

(√2)^3 + 3*(√2)^2*(√3) + 3*(√2)*(√3)^2 + (√3)^3.

Calculating each term:

First term: (√2)^3 = (√2)^2 * √2 = 2 * √2 = 2√2.

Second term: 3*(√2)^2*(√3) = 3*(2)*(√3) = 6√3.

Third term: 3*(√2)*(√3)^2 = 3*(√2)*(3) = 9√2.

Fourth term: (√3)^3 = (√3)^2 * √3 = 3 * √3 = 3√3.

Adding them all: 2√2 + 6√3 + 9√2 + 3√3 = (2√2 + 9√2) + (6√3 + 3√3) = 11√2 + 9√3.

So, no √6 term. That's consistent with both methods. Hmm. But the problem states that it can be written with a√6 term. So maybe the problem is incorrect? Or perhaps I need to check if there's a different expansion. Wait, let me check once again.

Alternatively, maybe the problem is (√2 + √3 + √6)^3? But the question says (√2 + √3)^3. Let me check the problem again.

Wait, the problem is definitely as written: (√2 + √3)^3. So that's correct. Then according to the expansion, it's 11√2 + 9√3, so a = 11, b = 9, c = 0. Therefore, a + b + c = 11 + 9 + 0 = 20. But the problem says that the expression can be written in the form a√2 + b√3 + c√6, so if c is zero, then technically it's a√2 + b√3 + 0√6. So maybe the answer is 20. But the problem is presented as if there's a non-zero c. Hmm. Alternatively, maybe I made a mistake in the expansion.

Wait, let me try expanding again, step by step, to make sure.

First, compute (√2 + √3)^3. Let's compute step by step:

First, compute (√2 + √3)^2:

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

Then multiply by (√2 + √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).

Now, √6*√2 = √(6*2) = √12 = 2√3.

Similarly, √6*√3 = √(6*3) = √18 = 3√2.

So substituting back:

5√2 + 5√3 + 2*(2√3) + 2*(3√2) = 5√2 + 5√3 + 4√3 + 6√2.

Combine like terms:

5√2 + 6√2 = 11√2

5√3 + 4√3 = 9√3

So total is 11√2 + 9√3. So again, no √6 term. Therefore, c = 0. Therefore, a + b + c = 11 + 9 + 0 = 20.

But the problem statement says "can be written in the form a√2 + b√3 + c√6". If c is zero, then that's acceptable. So the answer is 20.

Wait, but maybe the problem is expecting a different answer because there's a mistake in the problem statement. Alternatively, maybe I need to check once again.

Alternatively, perhaps the original problem is (√2 + √3 + √6)^3, but that's not what's given. Let me check the problem again.

No, the problem is definitely (√2 + √3)^3. Hmm. So unless there's a miscalculation, the answer should be 20. But let me check once again.

Alternatively, perhaps the problem expects the answer in terms of the minimal expression, but in this case, the expansion indeed does not have a √6 term. Therefore, c is 0. Therefore, the answer is 20.

Alternatively, maybe there's a mistake in the problem's statement. But since the user provided the problem as is, I should proceed with the calculation. Therefore, the answer is 20.
Therefore, the final answer is $\boxed{20}$.
</think>