Evalute: `(sqrt(3)+sqrt(2))^(6)- (sqrt(3)-sqrt(2))^(6)`

Explanation

To evaluate the expression \((\sqrt{3} + \sqrt{2})^6 - (\sqrt{3} - \sqrt{2})^6\), we can use the Binomial Theorem and some algebraic identities.

### Step-by-Step Solution:

1. **Identify the Terms**:
Let \( a = \sqrt{3} \) and \( b = \sqrt{2} \). We need to evaluate \( (a + b)^6 - (a - b)^6 \).

2. **Apply the Binomial Theorem**:
According to the Binomial Theorem:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
\[ (a - b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} (-b)^k \]

For \( n = 6 \):
\[ (a + b)^6 = \sum_{k=0}^{6} \binom{6}{k} a^{6-k} b^k \]
\[ (a - b)^6 = \sum_{k=0}^{6} \binom{6}{k} a^{6-k} (-b)^k \]

3. **Subtract the Two Expansions**:
When we subtract these two expansions:
\[ (a + b)^6 - (a - b)^6 = \sum_{k=0}^{6} \binom{6}{k} a^{6-k} b^k - \sum_{k=0}^{6} \binom{6}{k} a^{6-k} (-b)^k \]
Notice that when \( k \) is even, the terms cancel out, and when \( k \) is odd, they double:
\[ = 2 \sum_{k \text{ odd}} \binom{6}{k} a^{6-k} b^k \]

4. **Calculate the Odd Terms**:
The odd values of \( k \) from 0 to 6 are 1, 3, and 5:
- For \( k = 1 \):
\[ 2 \binom{6}{1} a^{5} b^{1} = 2 \cdot 6 \cdot (\sqrt{3})^5 \cdot \sqrt{2} = 12 \cdot 3^{2.5} \cdot \sqrt{2} = 12 \cdot 9\sqrt{3} \cdot \sqrt{2} = 108\sqrt{6} \]
- For \( k = 3 \):
\[ 2 \binom{6}{3} a^{3} b^{3} = 2 \cdot 20 \cdot (\sqrt{3})^3 \cdot (\sqrt{2})^3 = 40 \cdot 3^{1.5} \cdot 2^{1.5} = 40 \cdot 3\sqrt{3} \cdot 2\sqrt{2} = 240\sqrt{6} \]
- For \( k = 5 \):
\[ 2 \binom{6}{5} a^{1} b^{5} = 2 \cdot 6 \cdot \sqrt{3} \cdot (\sqrt{2})^5 = 12 \cdot \sqrt{3} \cdot 4\sqrt{2} = 48\sqrt{6} \]

5. **Combine the Results**:
Now, we add these contributions:
\[ 108\sqrt{6} + 240\sqrt{6} + 48\sqrt{6} = 396\sqrt{6} \]

### Final Answer:
\[ (\sqrt{3} + \sqrt{2})^6 - (\sqrt{3} - \sqrt{2})^6 = 396\sqrt{6} \]