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 $\sqrt{2} \times \sqrt{2} \times \sqrt{2}$. Let me compute that. $\sqrt{2} \times \sqrt{2} = 2$, so then multiplying by another $\sqrt{2}$ gives $2\sqrt{2}$. So the first term is $2\sqrt{2}$.

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

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

Fourth term: $(\sqrt{3})^3$. Similar to the first term. $\sqrt{3} \times \sqrt{3} = 3$, then multiply by another $\sqrt{3}$ gives $3\sqrt{3}$. So the fourth term is $3\sqrt{3}$.

Now, let me add up all the terms:

First term: $2\sqrt{2}$

Second term: $6\sqrt{3}$

Third term: $9\sqrt{2}$

Fourth term: $3\sqrt{3}$

So combining like terms. Let me group the $\sqrt{2}$ terms and the $\sqrt{3}$ terms.

For $\sqrt{2}$: $2\sqrt{2} + 9\sqrt{2} = (2 + 9)\sqrt{2} = 11\sqrt{2}$

For $\sqrt{3}$: $6\sqrt{3} + 3\sqrt{3} = (6 + 3)\sqrt{3} = 9\sqrt{3}$

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

Wait, let me check my expansion again. Maybe I missed a term? Let me go back through the expansion step by step.

Original expansion:

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

Yes, that's correct. So each term is accounted for. Let me check each term again.

First term: $(\sqrt{2})^3 = 2\sqrt{2}$. Correct.

Second term: $3(\sqrt{2})^2(\sqrt{3}) = 3*2*\sqrt{3} = 6\sqrt{3}$. Correct.

Third term: $3(\sqrt{2})(\sqrt{3})^2 = 3*\sqrt{2}*3 = 9\sqrt{2}$. Correct.

Fourth term: $(\sqrt{3})^3 = 3\sqrt{3}$. Correct.

So adding them up:

$\sqrt{2}$ terms: 2 + 9 = 11

$\sqrt{3}$ terms: 6 + 3 = 9

So total expression is $11\sqrt{2} + 9\sqrt{3}$. But the problem says it should be in the form with a $\sqrt{6}$ term. So where is the mistake here?

Wait, maybe I need to check if there's a cross term that I missed? Wait, no. The expansion of $(a + b)^3$ is indeed $a^3 + 3a^2b + 3ab^2 + b^3$. So unless there's a miscalculation here, but I don't see it.

Alternatively, maybe the problem is expecting me to consider that when expanding $(\sqrt{2} + \sqrt{3})^3$, there might be a $\sqrt{6}$ term? But according to my calculation, there isn't. Wait, let me try expanding it in another way, maybe multiplying step by step.

Let me compute $(\sqrt{2} + \sqrt{3})^3$ as $(\sqrt{2} + \sqrt{3})^2 \times (\sqrt{2} + \sqrt{3})$.

First, compute $(\sqrt{2} + \sqrt{3})^2$. Let me do that:

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

Okay, so that's $5 + 2\sqrt{6}$. Then, multiply this by $(\sqrt{2} + \sqrt{3})$:

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

Now, let me compute each term:

First term: $5\sqrt{2}$

Second term: $5\sqrt{3}$

Third term: $2\sqrt{6}\sqrt{2}$. Let me simplify $\sqrt{6}\sqrt{2}$. Since $\sqrt{6} \times \sqrt{2} = \sqrt{12} = 2\sqrt{3}$. Therefore, this term is $2 \times 2\sqrt{3} = 4\sqrt{3}$.

Fourth term: $2\sqrt{6}\sqrt{3}$. Similarly, $\sqrt{6}\sqrt{3} = \sqrt{18} = 3\sqrt{2}$. Therefore, this term is $2 \times 3\sqrt{2} = 6\sqrt{2}$.

So now, combining all terms:

First term: $5\sqrt{2}$

Second term: $5\sqrt{3}$

Third term: $4\sqrt{3}$

Fourth term: $6\sqrt{2}$

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, the same result: $11\sqrt{2} + 9\sqrt{3}$. But the problem says it should be in the form with a $\sqrt{6}$ term. So why is there no $\sqrt{6}$ term here?

Wait, maybe the problem is wrong? Or maybe I misread it? Let me check again.

The problem 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 $a+b+c$?"

But according to both methods, expanding directly and expanding step by step, there is no $\sqrt{6}$ term. So either the problem is incorrect, or I made a mistake. Wait, but maybe I need to check my calculations again?

Wait, let me check the second method again. When I multiplied $(5 + 2\sqrt{6})(\sqrt{2} + \sqrt{3})$, let me verify each term again.

First term: 5 * sqrt(2) = 5√2. Correct.

Second term: 5 * sqrt(3) = 5√3. Correct.

Third term: 2√6 * sqrt(2). Let me compute √6 * √2. √(6*2) = √12 = 2√3. So 2 * 2√3 = 4√3. Correct.

Fourth term: 2√6 * sqrt(3). √6 * √3 = √(6*3) = √18 = 3√2. Then 2 * 3√2 = 6√2. Correct.

So combining terms gives 5√2 + 6√2 = 11√2 and 5√3 + 4√3 = 9√3. So no √6 term. So why does the problem say it can be written with a √6 term?

Wait, maybe there's a mistake in the problem? Or maybe I need to consider that maybe there is a √6 term, but it cancels out? But in both methods, there is no √6 term. Wait, unless I made a mistake in the expansion.

Alternatively, maybe the problem is expecting me to consider that when expanding $(\sqrt{2} + \sqrt{3})^3$, there is a cross term that results in √6? But according to both methods, there isn't. Wait, let me try expanding using another approach.

Alternatively, maybe I can use the binomial theorem again but check for errors. Let me write out the 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: $\binom{3}{0} (\sqrt{2})^3 (\sqrt{3})^0 = 1 * 2\sqrt{2} * 1 = 2\sqrt{2}$

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

Adding them up: 2√2 + 6√3 + 9√2 + 3√3 = (2 + 9)√2 + (6 + 3)√3 = 11√2 + 9√3. Same result. So no √6 term. Therefore, according to this, the expression is 11√2 + 9√3 + 0√6. Therefore, a = 11, b = 9, c = 0. Then a + b + c = 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? The problem says "where a, b, and c are integers". So zero is an integer. Therefore, maybe the answer is 20?

But why does the problem mention the √6 term? Maybe it's a trick question? Or maybe I made a mistake in thinking that there is no √6 term? Wait, let me check with decimal approximations.

Let me compute the numerical value of $(\sqrt{2} + \sqrt{3})^3$ and also compute 11√2 + 9√3 and see if they match. If they do, then maybe the answer is indeed 20.

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

First, 3^3 = 27. Let me compute 3.1462^3:

First, compute 3.1462 * 3.1462:

Approximate 3.1462^2:

Let me compute 3.1462 * 3.1462:

Well, 3 * 3 = 9, 3 * 0.1462 = 0.4386, 0.1462 * 3 = 0.4386, and 0.1462^2 ≈ 0.0214. So total is approximately 9 + 0.4386 + 0.4386 + 0.0214 ≈ 9 + 0.8772 + 0.0214 ≈ 9.8986. So approximately 9.8986. Then multiply by 3.1462:

So 9.8986 * 3.1462 ≈ Let me approximate this. Let me take 10 * 3.1462 = 31.462, subtract 0.1014 * 3.1462 ≈ 0.3188. So approximately 31.462 - 0.3188 ≈ 31.1432.

Now, compute 11√2 + 9√3:

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 approximately 31.1432. So that seems to check out. Therefore, the value of $(\sqrt{2} + \sqrt{3})^3$ is indeed approximately 31.14, which matches with 11√2 + 9√3. Therefore, there is no √6 term. Therefore, c = 0. Therefore, a + b + c = 11 + 9 + 0 = 20.

But the problem says "can be written in the form a√2 + b√3 + c√6". So maybe the problem is designed to have c = 0? Maybe the question is testing whether we can recognize that even though the form includes a √6 term, it's possible for c to be zero. Therefore, the answer is 20. But maybe I need to check again if there's a different approach where √6 term appears?

Alternatively, perhaps I made a mistake in the problem statement? Let me check again. The problem 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 $a+b+c$?"

Alternatively, maybe there's a different expansion? Wait, let me think. Suppose I consider that when expanding $(\sqrt{2} + \sqrt{3})^3$, maybe there is a term with √6? But according to all calculations, there isn't. Wait, unless I made a mistake in the multiplication steps. Let me check once again.

Wait, when I expanded $(\sqrt{2} + \sqrt{3})^3$ as $(\sqrt{2} + \sqrt{3})^2 \times (\sqrt{2} + \sqrt{3})$, I first calculated $(\sqrt{2} + \sqrt{3})^2 = 5 + 2\sqrt{6}$. Then multiplied by $\sqrt{2} + \sqrt{3}$. Let me check that 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 up: 5√2 + 6√2 = 11√2, and 5√3 + 4√3 = 9√3. So again, same result. Therefore, no √6 term. Therefore, c = 0. Therefore, the answer is 11 + 9 + 0 = 20.

But maybe the problem had a typo? For example, if it was $(\sqrt{2} + \sqrt{3})^2$, then it would be 5 + 2√6, which would have a √6 term. But the problem says cubed. Alternatively, maybe the original problem was different? But assuming the problem is correct, then the answer is 20.

Alternatively, maybe I need to consider that when expanding $(\sqrt{2} + \sqrt{3})^3$, there is a √6 term that somehow combines with others? But according to the calculations, there isn't. Wait, unless I made a mistake in the binomial expansion. Let me check once again.

Wait, let me think of another approach. Let me compute $(\sqrt{2} + \sqrt{3})^3$ by multiplying $(\sqrt{2} + \sqrt{3})$ three times. Let me do that 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 $(\sqrt{2} + \sqrt{3})$:

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

As before, √6√2 = √12 = 2√3, so 2*2√3 = 4√3.

Similarly, √6√3 = √18 = 3√2, so 2*3√2 = 6√2.

So total is 5√2 + 6√2 = 11√2, and 5√3 + 4√3 = 9√3. So again, same result. Therefore, no √6 term. Therefore, c = 0.

Therefore, the answer is 11 + 9 + 0 = 20. Therefore, maybe the problem is correct, and c is allowed to be zero. Therefore, the answer is 20.

Alternatively, maybe the problem is expecting me to consider that there is a √6 term, but I need to combine terms differently? Wait, but in the expression 11√2 + 9√3, there is no √6 term. Therefore, unless there is a miscalculation, which I have checked multiple times, I think that the answer is 20.

Alternatively, maybe the problem is from a source where the answer is different? Wait, but I need to trust my calculations here. Let me check with another method. Let me compute $(\sqrt{2} + \sqrt{3})^3$ numerically and compare with 11√2 + 9√3 and also with 11√2 + 9√3 + 0√6.

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

First, compute $\sqrt{2} ≈ 1.41421356$, $\sqrt{3} ≈ 1.73205081$.

Adding them: 1.41421356 + 1.73205081 ≈ 3.14626437.

Cubing that: 3.14626437^3.

Let me compute 3.14626437^3:

First, compute 3.14626437 * 3.14626437:

Let me compute 3.14626437 * 3.14626437:

Let me approximate this as follows:

Let me take 3.14626437 ≈ 3.146264

Let me compute 3.146264^2:

First, 3^2 = 9.

Then, 0.146264^2 ≈ 0.02139.

Then, cross terms: 2 * 3 * 0.146264 ≈ 0.877584.

So total ≈ 9 + 0.877584 + 0.02139 ≈ 9.898974.

So approximately 9.898974.

Now, multiply by 3.146264:

So 9.898974 * 3.146264 ≈ Let me compute this:

First, 10 * 3.146264 = 31.46264

Subtract 0.101026 * 3.146264 ≈ 0.101026 * 3.146264 ≈ approx 0.3178.

Therefore, approx 31.46264 - 0.3178 ≈ 31.1448.

Now, compute 11√2 + 9√3:

11 * 1.41421356 ≈ 15.55634916

9 * 1.73205081 ≈ 15.58845729

Adding them: 15.55634916 + 15.58845729 ≈ 31.14480645

Which matches the previous value of approximately 31.1448. Therefore, this confirms that $(\sqrt{2} + \sqrt{3})^3 ≈ 31.1448$, which is equal to 11√2 + 9√3. Therefore, there is no √6 term. Therefore, c = 0. Therefore, the answer is 11 + 9 + 0 = 20.

Therefore, maybe the problem is correct, and the answer is 20. Therefore, I think that the answer is 20. However, since the problem mentions the form with a √6 term, maybe there's a different interpretation?

Wait, let me think again. Suppose that the problem is written as $(\sqrt{2} + \sqrt{3})^3$, but maybe there's a different expansion? Wait, no. Alternatively, maybe the problem is in another form? For example, if it was $(\sqrt{2} + \sqrt{3})^3 + something$, but no, the problem says just $(\sqrt{2} + \sqrt{3})^3$.

Alternatively, maybe the problem is expecting me to consider that when expanding, there is a √6 term, but I need to combine terms? Wait, but in the expansion, there is no √6 term. Wait, unless I made a mistake in the expansion. Wait, let me think again.

Wait, let me try expanding $(\sqrt{2} + \sqrt{3})^3$ using the distributive property step by step:

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

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

Now, multiply this by the third factor $(\sqrt{2} + \sqrt{3})$:

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

As before, √6√2 = √12 = 2√3, so 2*2√3 = 4√3.

Similarly, √6√3 = √18 = 3√2, so 2*3√2 = 6√2.

Therefore, adding up:

5√2 + 6√2 = 11√2

5√3 + 4√3 = 9√3

So again, no √6 term. Therefore, the conclusion is that c = 0. Therefore, the answer is 20.

Alternatively, maybe the problem had a different exponent? For example, if it was squared, then we have 5 + 2√6, which would have a √6 term. But the problem says cubed. Therefore, unless there's a typo, the answer is 20.

Alternatively, maybe the problem is from a source where the answer is different? But given that my calculations check out with numerical approximations, I think that 20 is the correct answer.

Therefore, I think the answer is 20, so a + b + c = 11 + 9 + 0 = 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 value of $a + b + c$.

---

### Step 1: Expand the Expression

We use the binomial expansion for $(\sqrt{2} + \sqrt{3})^3$:

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

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:

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

So the expression becomes:

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

---

### Step 2: Match the Form

We are told the expression can be written in the form:

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

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 $\sqrt{2} \times \sqrt{2} \times \sqrt{2}$. Let me compute that. $\sqrt{2} \times \sqrt{2} = 2$, so then multiplying by another $\sqrt{2}$ gives $2\sqrt{2}$. So the first term is $2\sqrt{2}$.

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

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

Fourth term: $(\sqrt{3})^3$. Similar to the first term. $\sqrt{3} \times \sqrt{3} = 3$, then multiply by another $\sqrt{3}$ gives $3\sqrt{3}$. So the fourth term is $3\sqrt{3}$.

Now, let me add up all the terms:

First term: $2\sqrt{2}$

Second term: $6\sqrt{3}$

Third term: $9\sqrt{2}$

Fourth term: $3\sqrt{3}$

So combining like terms. Let me group the $\sqrt{2}$ terms and the $\sqrt{3}$ terms.

For $\sqrt{2}$: $2\sqrt{2} + 9\sqrt{2} = (2 + 9)\sqrt{2} = 11\sqrt{2}$

For $\sqrt{3}$: $6\sqrt{3} + 3\sqrt{3} = (6 + 3)\sqrt{3} = 9\sqrt{3}$

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

Wait, let me check my expansion again. Maybe I missed a term? Let me go back through the expansion step by step.

Original expansion:

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

Yes, that's correct. So each term is accounted for. Let me check each term again.

First term: $(\sqrt{2})^3 = 2\sqrt{2}$. Correct.

Second term: $3(\sqrt{2})^2(\sqrt{3}) = 3*2*\sqrt{3} = 6\sqrt{3}$. Correct.

Third term: $3(\sqrt{2})(\sqrt{3})^2 = 3*\sqrt{2}*3 = 9\sqrt{2}$. Correct.

Fourth term: $(\sqrt{3})^3 = 3\sqrt{3}$. Correct.

So adding them up:

$\sqrt{2}$ terms: 2 + 9 = 11

$\sqrt{3}$ terms: 6 + 3 = 9

So total expression is $11\sqrt{2} + 9\sqrt{3}$. But the problem says it should be in the form with a $\sqrt{6}$ term. So where is the mistake here?

Wait, maybe I need to check if there's a cross term that I missed? Wait, no. The expansion of $(a + b)^3$ is indeed $a^3 + 3a^2b + 3ab^2 + b^3$. So unless there's a miscalculation here, but I don't see it.

Alternatively, maybe the problem is expecting me to consider that when expanding $(\sqrt{2} + \sqrt{3})^3$, there might be a $\sqrt{6}$ term? But according to my calculation, there isn't. Wait, let me try expanding it in another way, maybe multiplying step by step.

Let me compute $(\sqrt{2} + \sqrt{3})^3$ as $(\sqrt{2} + \sqrt{3})^2 \times (\sqrt{2} + \sqrt{3})$.

First, compute $(\sqrt{2} + \sqrt{3})^2$. Let me do that:

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

Okay, so that's $5 + 2\sqrt{6}$. Then, multiply this by $(\sqrt{2} + \sqrt{3})$:

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

Now, let me compute each term:

First term: $5\sqrt{2}$

Second term: $5\sqrt{3}$

Third term: $2\sqrt{6}\sqrt{2}$. Let me simplify $\sqrt{6}\sqrt{2}$. Since $\sqrt{6} \times \sqrt{2} = \sqrt{12} = 2\sqrt{3}$. Therefore, this term is $2 \times 2\sqrt{3} = 4\sqrt{3}$.

Fourth term: $2\sqrt{6}\sqrt{3}$. Similarly, $\sqrt{6}\sqrt{3} = \sqrt{18} = 3\sqrt{2}$. Therefore, this term is $2 \times 3\sqrt{2} = 6\sqrt{2}$.

So now, combining all terms:

First term: $5\sqrt{2}$

Second term: $5\sqrt{3}$

Third term: $4\sqrt{3}$

Fourth term: $6\sqrt{2}$

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, the same result: $11\sqrt{2} + 9\sqrt{3}$. But the problem says it should be in the form with a $\sqrt{6}$ term. So why is there no $\sqrt{6}$ term here?

Wait, maybe the problem is wrong? Or maybe I misread it? Let me check again.

The problem 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 $a+b+c$?"

But according to both methods, expanding directly and expanding step by step, there is no $\sqrt{6}$ term. So either the problem is incorrect, or I made a mistake. Wait, but maybe I need to check my calculations again?

Wait, let me check the second method again. When I multiplied $(5 + 2\sqrt{6})(\sqrt{2} + \sqrt{3})$, let me verify each term again.

First term: 5 * sqrt(2) = 5√2. Correct.

Second term: 5 * sqrt(3) = 5√3. Correct.

Third term: 2√6 * sqrt(2). Let me compute √6 * √2. √(6*2) = √12 = 2√3. So 2 * 2√3 = 4√3. Correct.

Fourth term: 2√6 * sqrt(3). √6 * √3 = √(6*3) = √18 = 3√2. Then 2 * 3√2 = 6√2. Correct.

So combining terms gives 5√2 + 6√2 = 11√2 and 5√3 + 4√3 = 9√3. So no √6 term. So why does the problem say it can be written with a √6 term?

Wait, maybe there's a mistake in the problem? Or maybe I need to consider that maybe there is a √6 term, but it cancels out? But in both methods, there is no √6 term. Wait, unless I made a mistake in the expansion.

Alternatively, maybe the problem is expecting me to consider that when expanding $(\sqrt{2} + \sqrt{3})^3$, there is a cross term that results in √6? But according to both methods, there isn't. Wait, let me try expanding using another approach.

Alternatively, maybe I can use the binomial theorem again but check for errors. Let me write out the 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: $\binom{3}{0} (\sqrt{2})^3 (\sqrt{3})^0 = 1 * 2\sqrt{2} * 1 = 2\sqrt{2}$

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

Adding them up: 2√2 + 6√3 + 9√2 + 3√3 = (2 + 9)√2 + (6 + 3)√3 = 11√2 + 9√3. Same result. So no √6 term. Therefore, according to this, the expression is 11√2 + 9√3 + 0√6. Therefore, a = 11, b = 9, c = 0. Then a + b + c = 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? The problem says "where a, b, and c are integers". So zero is an integer. Therefore, maybe the answer is 20?

But why does the problem mention the √6 term? Maybe it's a trick question? Or maybe I made a mistake in thinking that there is no √6 term? Wait, let me check with decimal approximations.

Let me compute the numerical value of $(\sqrt{2} + \sqrt{3})^3$ and also compute 11√2 + 9√3 and see if they match. If they do, then maybe the answer is indeed 20.

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

First, 3^3 = 27. Let me compute 3.1462^3:

First, compute 3.1462 * 3.1462:

Approximate 3.1462^2:

Let me compute 3.1462 * 3.1462:

Well, 3 * 3 = 9, 3 * 0.1462 = 0.4386, 0.1462 * 3 = 0.4386, and 0.1462^2 ≈ 0.0214. So total is approximately 9 + 0.4386 + 0.4386 + 0.0214 ≈ 9 + 0.8772 + 0.0214 ≈ 9.8986. So approximately 9.8986. Then multiply by 3.1462:

So 9.8986 * 3.1462 ≈ Let me approximate this. Let me take 10 * 3.1462 = 31.462, subtract 0.1014 * 3.1462 ≈ 0.3188. So approximately 31.462 - 0.3188 ≈ 31.1432.

Now, compute 11√2 + 9√3:

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 approximately 31.1432. So that seems to check out. Therefore, the value of $(\sqrt{2} + \sqrt{3})^3$ is indeed approximately 31.14, which matches with 11√2 + 9√3. Therefore, there is no √6 term. Therefore, c = 0. Therefore, a + b + c = 11 + 9 + 0 = 20.

But the problem says "can be written in the form a√2 + b√3 + c√6". So maybe the problem is designed to have c = 0? Maybe the question is testing whether we can recognize that even though the form includes a √6 term, it's possible for c to be zero. Therefore, the answer is 20. But maybe I need to check again if there's a different approach where √6 term appears?

Alternatively, perhaps I made a mistake in the problem statement? Let me check again. The problem 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 $a+b+c$?"

Alternatively, maybe there's a different expansion? Wait, let me think. Suppose I consider that when expanding $(\sqrt{2} + \sqrt{3})^3$, maybe there is a term with √6? But according to all calculations, there isn't. Wait, unless I made a mistake in the multiplication steps. Let me check once again.

Wait, when I expanded $(\sqrt{2} + \sqrt{3})^3$ as $(\sqrt{2} + \sqrt{3})^2 \times (\sqrt{2} + \sqrt{3})$, I first calculated $(\sqrt{2} + \sqrt{3})^2 = 5 + 2\sqrt{6}$. Then multiplied by $\sqrt{2} + \sqrt{3}$. Let me check that 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 up: 5√2 + 6√2 = 11√2, and 5√3 + 4√3 = 9√3. So again, same result. Therefore, no √6 term. Therefore, c = 0. Therefore, the answer is 11 + 9 + 0 = 20.

But maybe the problem had a typo? For example, if it was $(\sqrt{2} + \sqrt{3})^2$, then it would be 5 + 2√6, which would have a √6 term. But the problem says cubed. Alternatively, maybe the original problem was different? But assuming the problem is correct, then the answer is 20.

Alternatively, maybe I need to consider that when expanding $(\sqrt{2} + \sqrt{3})^3$, there is a √6 term that somehow combines with others? But according to the calculations, there isn't. Wait, unless I made a mistake in the binomial expansion. Let me check once again.

Wait, let me think of another approach. Let me compute $(\sqrt{2} + \sqrt{3})^3$ by multiplying $(\sqrt{2} + \sqrt{3})$ three times. Let me do that 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 $(\sqrt{2} + \sqrt{3})$:

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

As before, √6√2 = √12 = 2√3, so 2*2√3 = 4√3.

Similarly, √6√3 = √18 = 3√2, so 2*3√2 = 6√2.

So total is 5√2 + 6√2 = 11√2, and 5√3 + 4√3 = 9√3. So again, same result. Therefore, no √6 term. Therefore, c = 0.

Therefore, the answer is 11 + 9 + 0 = 20. Therefore, maybe the problem is correct, and c is allowed to be zero. Therefore, the answer is 20.

Alternatively, maybe the problem is expecting me to consider that there is a √6 term, but I need to combine terms differently? Wait, but in the expression 11√2 + 9√3, there is no √6 term. Therefore, unless there is a miscalculation, which I have checked multiple times, I think that the answer is 20.

Alternatively, maybe the problem is from a source where the answer is different? Wait, but I need to trust my calculations here. Let me check with another method. Let me compute $(\sqrt{2} + \sqrt{3})^3$ numerically and compare with 11√2 + 9√3 and also with 11√2 + 9√3 + 0√6.

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

First, compute $\sqrt{2} ≈ 1.41421356$, $\sqrt{3} ≈ 1.73205081$.

Adding them: 1.41421356 + 1.73205081 ≈ 3.14626437.

Cubing that: 3.14626437^3.

Let me compute 3.14626437^3:

First, compute 3.14626437 * 3.14626437:

Let me compute 3.14626437 * 3.14626437:

Let me approximate this as follows:

Let me take 3.14626437 ≈ 3.146264

Let me compute 3.146264^2:

First, 3^2 = 9.

Then, 0.146264^2 ≈ 0.02139.

Then, cross terms: 2 * 3 * 0.146264 ≈ 0.877584.

So total ≈ 9 + 0.877584 + 0.02139 ≈ 9.898974.

So approximately 9.898974.

Now, multiply by 3.146264:

So 9.898974 * 3.146264 ≈ Let me compute this:

First, 10 * 3.146264 = 31.46264

Subtract 0.101026 * 3.146264 ≈ 0.101026 * 3.146264 ≈ approx 0.3178.

Therefore, approx 31.46264 - 0.3178 ≈ 31.1448.

Now, compute 11√2 + 9√3:

11 * 1.41421356 ≈ 15.55634916

9 * 1.73205081 ≈ 15.58845729

Adding them: 15.55634916 + 15.58845729 ≈ 31.14480645

Which matches the previous value of approximately 31.1448. Therefore, this confirms that $(\sqrt{2} + \sqrt{3})^3 ≈ 31.1448$, which is equal to 11√2 + 9√3. Therefore, there is no √6 term. Therefore, c = 0. Therefore, the answer is 11 + 9 + 0 = 20.

Therefore, maybe the problem is correct, and the answer is 20. Therefore, I think that the answer is 20. However, since the problem mentions the form with a √6 term, maybe there's a different interpretation?

Wait, let me think again. Suppose that the problem is written as $(\sqrt{2} + \sqrt{3})^3$, but maybe there's a different expansion? Wait, no. Alternatively, maybe the problem is in another form? For example, if it was $(\sqrt{2} + \sqrt{3})^3 + something$, but no, the problem says just $(\sqrt{2} + \sqrt{3})^3$.

Alternatively, maybe the problem is expecting me to consider that when expanding, there is a √6 term, but I need to combine terms? Wait, but in the expansion, there is no √6 term. Wait, unless I made a mistake in the expansion. Wait, let me think again.

Wait, let me try expanding $(\sqrt{2} + \sqrt{3})^3$ using the distributive property step by step:

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

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

Now, multiply this by the third factor $(\sqrt{2} + \sqrt{3})$:

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

As before, √6√2 = √12 = 2√3, so 2*2√3 = 4√3.

Similarly, √6√3 = √18 = 3√2, so 2*3√2 = 6√2.

Therefore, adding up:

5√2 + 6√2 = 11√2

5√3 + 4√3 = 9√3

So again, no √6 term. Therefore, the conclusion is that c = 0. Therefore, the answer is 20.

Alternatively, maybe the problem had a different exponent? For example, if it was squared, then we have 5 + 2√6, which would have a √6 term. But the problem says cubed. Therefore, unless there's a typo, the answer is 20.

Alternatively, maybe the problem is from a source where the answer is different? But given that my calculations check out with numerical approximations, I think that 20 is the correct answer.

Therefore, I think the answer is 20, so a + b + c = 11 + 9 + 0 = 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 value of $a + b + c$.

---

### Step 1: Expand the Expression

We use the binomial expansion for $(\sqrt{2} + \sqrt{3})^3$:

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

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:

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

So the expression becomes:

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

---

### Step 2: Match the Form

We are told the expression can be written in the form:

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

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 $a + b + c = \boxed{20}$.