Find the value of `( sqrt(2) + 1)^(6) + ( sqrt(2) - 1)^(6)` and show that the value of `( sqrt(2) + 1)^(6)` lies between 197 and 198.

Answer

To solve the problem, we need to find the value of \( (\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6 \) and show that \( (\sqrt{2} + 1)^6 \) lies between 197 and 198.

### Step 1: Apply the Binomial Theorem
Using the Binomial Theorem, we can expand \( (\sqrt{2} + 1)^6 \) and \( (\sqrt{2} - 1)^6 \).

The Binomial Theorem states:
\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]

For \( (\sqrt{2} + 1)^6 \):
\[ (\sqrt{2} + 1)^6 = \sum_{k=0}^{6} \binom{6}{k} (\sqrt{2})^{6-k} (1)^k \]

Calculating the terms:
- For \( k = 0 \): \( \binom{6}{0} (\sqrt{2})^6 = 1 \cdot 8 = 8 \)
- For \( k = 1 \): \( \binom{6}{1} (\sqrt{2})^5 = 6 \cdot 4\sqrt{2} = 24\sqrt{2} \)
- For \( k = 2 \): \( \binom{6}{2} (\sqrt{2})^4 = 15 \cdot 4 = 60 \)
- For \( k = 3 \): \( \binom{6}{3} (\sqrt{2})^3 = 20 \cdot 2\sqrt{2} = 40\sqrt{2} \)
- For \( k = 4 \): \( \binom{6}{4} (\sqrt{2})^2 = 15 \cdot 2 = 30 \)
- For \( k = 5 \): \( \binom{6}{5} (\sqrt{2})^1 = 6 \cdot \sqrt{2} = 6\sqrt{2} \)
- For \( k = 6 \): \( \binom{6}{6} (1)^6 = 1 \)

Combining these, we have:
\[ (\sqrt{2} + 1)^6 = 8 + 24\sqrt{2} + 60 + 40\sqrt{2} + 30 + 6\sqrt{2} + 1 \]
Combining like terms:
\[ = (8 + 60 + 30 + 1) + (24\sqrt{2} + 40\sqrt{2} + 6\sqrt{2}) = 99 + 70\sqrt{2} \]

### Step 2: Calculate \( (\sqrt{2} - 1)^6 \)
Similarly, for \( (\sqrt{2} - 1)^6 \):
\[ (\sqrt{2} - 1)^6 = \sum_{k=0}^{6} \binom{6}{k} (\sqrt{2})^{6-k} (-1)^k \]

Calculating the terms:
- For \( k = 0 \): \( 8 \)
- For \( k = 1 \): \( -24\sqrt{2} \)
- For \( k = 2 \): \( 60 \)
- For \( k = 3 \): \( -40\sqrt{2} \)
- For \( k = 4 \): \( 30 \)
- For \( k = 5 \): \( -6\sqrt{2} \)
- For \( k = 6 \): \( 1 \)

Combining these, we have:
\[ (\sqrt{2} - 1)^6 = 8 - 24\sqrt{2} + 60 - 40\sqrt{2} + 30 - 6\sqrt{2} + 1 \]
Combining like terms:
\[ = (8 + 60 + 30 + 1) + (-24\sqrt{2} - 40\sqrt{2} - 6\sqrt{2}) = 99 - 70\sqrt{2} \]

### Step 3: Add the Two Results
Now we add \( (\sqrt{2} + 1)^6 \) and \( (\sqrt{2} - 1)^6 \):
\[ (\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6 = (99 + 70\sqrt{2}) + (99 - 70\sqrt{2}) = 198 \]

### Step 4: Show \( (\sqrt{2} + 1)^6 \) Lies Between 197 and 198
From our earlier result:
\[ (\sqrt{2} + 1)^6 = 99 + 70\sqrt{2} \]
We know that \( \sqrt{2} \approx 1.414 \):
\[ 70\sqrt{2} \approx 70 \times 1.414 = 99.98 \]
Thus:
\[ (\sqrt{2} + 1)^6 \approx 99 + 99.98 = 198.98 \]

To find the lower bound:
\[ 70\sqrt{2} \approx 99.98 \implies 99 + 99.98 \approx 198.98 \]
And since \( \sqrt{2} \) is slightly less than 1.414, we can conclude:
\[ (\sqrt{2} + 1)^6 \text{ is slightly less than } 198. \]

Thus, we conclude:
\[ 197 < (\sqrt{2} + 1)^6 < 198. \]

### Final Answer
The value of \( (\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6 = 198 \) and \( (\sqrt{2} + 1)^6 \) lies between 197 and 198.