`(sqrt(2) + 1)^(6) - (sqrt(2) -1)^(6)` is equal to

Options

A. 101
B. `70sqrt(2)`
C. `140 sqrt(2)`
D. `120sqrt(2)`

Answer

To solve the expression \((\sqrt{2} + 1)^{6} - (\sqrt{2} - 1)^{6}\), we will use the Binomial Theorem and some algebraic manipulation.

### Step-by-Step Solution:

1. **Rewrite the Expression**:
We start with the expression:
\[ (\sqrt{2} + 1)^{6} - (\sqrt{2} - 1)^{6} \]

2. **Use the Binomial Theorem**:
According to the Binomial Theorem, we can expand both terms:
\[ (\sqrt{2} + 1)^{6} = \sum_{k=0}^{6} \binom{6}{k} (\sqrt{2})^{6-k} (1)^{k} \]
\[ (\sqrt{2} - 1)^{6} = \sum_{k=0}^{6} \binom{6}{k} (\sqrt{2})^{6-k} (-1)^{k} \]

3. **Combine the Expansions**:
When we subtract the two expansions, we notice that the terms with even \(k\) will cancel out:
\[ (\sqrt{2} + 1)^{6} - (\sqrt{2} - 1)^{6} = 2 \sum_{k \text{ odd}} \binom{6}{k} (\sqrt{2})^{6-k} \]

4. **Identify the Odd Terms**:
The odd values of \(k\) from \(0\) to \(6\) are \(1, 3, 5\). We calculate these terms:
- For \(k = 1\):
\[ \binom{6}{1} (\sqrt{2})^{5} = 6 \cdot 2^{5/2} = 6 \cdot 4\sqrt{2} = 24\sqrt{2} \]
- For \(k = 3\):
\[ \binom{6}{3} (\sqrt{2})^{3} = 20 \cdot 2^{3/2} = 20 \cdot 2\sqrt{2} = 40\sqrt{2} \]
- For \(k = 5\):
\[ \binom{6}{5} (\sqrt{2})^{1} = 6 \cdot \sqrt{2} = 6\sqrt{2} \]

5. **Sum the Odd Terms**:
Now we sum these contributions:
\[ 2 \left( 24\sqrt{2} + 40\sqrt{2} + 6\sqrt{2} \right) = 2 \cdot 70\sqrt{2} = 140\sqrt{2} \]

6. **Final Result**:
Therefore, the final result is:
\[ (\sqrt{2} + 1)^{6} - (\sqrt{2} - 1)^{6} = 140\sqrt{2} \]