Evalute: `(sqrt(2)+1)^(6) + ( sqrt(2) - 1)^(6)`

Explanation

To evaluate the expression \((\sqrt{2}+1)^{6} + (\sqrt{2}-1)^{6}\), we can use the Binomial Theorem and some algebraic manipulations. Let's break it down step by step.

### Step 1: Rewrite the Expression
We can express the given terms as:
\[ (\sqrt{2}+1)^{6} + (\sqrt{2}-1)^{6} \]
This can be rewritten using the property of exponents:
\[ = \left((\sqrt{2}+1)^{2}\right)^{3} + \left((\sqrt{2}-1)^{2}\right)^{3} \]

### Step 2: Calculate \((\sqrt{2}+1)^{2}\) and \((\sqrt{2}-1)^{2}\)
Now we calculate each square:
\[ (\sqrt{2}+1)^{2} = (\sqrt{2})^{2} + 2 \cdot \sqrt{2} \cdot 1 + 1^{2} = 2 + 2\sqrt{2} + 1 = 3 + 2\sqrt{2} \]
\[ (\sqrt{2}-1)^{2} = (\sqrt{2})^{2} - 2 \cdot \sqrt{2} \cdot 1 + 1^{2} = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2} \]

### Step 3: Substitute Back into the Expression
Now substitute these results back into the expression:
\[ = (3 + 2\sqrt{2})^{3} + (3 - 2\sqrt{2})^{3} \]

### Step 4: Use the Sum of Cubes Formula
We can use the formula for the sum of cubes:
\[ a^{3} + b^{3} = (a+b)(a^{2} - ab + b^{2}) \]
Here, let \(a = 3 + 2\sqrt{2}\) and \(b = 3 - 2\sqrt{2}\).

First, calculate \(a + b\):
\[ a + b = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6 \]

### Step 5: Calculate \(a^{2}\) and \(b^{2}\)
Next, we calculate \(a^{2}\) and \(b^{2}\):
\[ a^{2} = (3 + 2\sqrt{2})^{2} = 9 + 12\sqrt{2} + 8 = 17 + 12\sqrt{2} \]
\[ b^{2} = (3 - 2\sqrt{2})^{2} = 9 - 12\sqrt{2} + 8 = 17 - 12\sqrt{2} \]

### Step 6: Calculate \(ab\)
Now calculate \(ab\):
\[ ab = (3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 9 - 8 = 1 \]

### Step 7: Substitute into the Sum of Cubes Formula
Now substitute these values back into the sum of cubes formula:
\[ a^{3} + b^{3} = (6)((17 + 12\sqrt{2}) + (17 - 12\sqrt{2}) - 1) \]
\[ = 6(34 - 1) = 6 \times 33 = 198 \]

### Final Answer
Thus, the value of \((\sqrt{2}+1)^{6} + (\sqrt{2}-1)^{6}\) is:
\[ \boxed{198} \]