The value of `{(sqrt(2) +1)^(5) + (sqrt(2) -1)^(5)}` ,is

Options

A. 58
B. `58sqrt(2)`
C. 42
D. `42sqrt(2)`

Explanation

To find the value of \((\sqrt{2} + 1)^{5} + (\sqrt{2} - 1)^{5}\), we can use the Binomial Theorem. The Binomial Theorem states that:

\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]

### Step 1: Apply the Binomial Theorem

We will apply the Binomial Theorem to both \((\sqrt{2} + 1)^{5}\) and \((\sqrt{2} - 1)^{5}\).

For \((\sqrt{2} + 1)^{5}\):

\[ (\sqrt{2} + 1)^{5} = \sum_{k=0}^{5} \binom{5}{k} (\sqrt{2})^{5-k} (1)^{k} \]

For \((\sqrt{2} - 1)^{5}\):

\[ (\sqrt{2} - 1)^{5} = \sum_{k=0}^{5} \binom{5}{k} (\sqrt{2})^{5-k} (-1)^{k} \]

### Step 2: Add the Two Expansions

Now, we add the two expansions:

\[ (\sqrt{2} + 1)^{5} + (\sqrt{2} - 1)^{5} = \sum_{k=0}^{5} \binom{5}{k} (\sqrt{2})^{5-k} (1)^{k} + \sum_{k=0}^{5} \binom{5}{k} (\sqrt{2})^{5-k} (-1)^{k} \]

### Step 3: Simplify the Expression

When we add these two expansions, the terms where \(k\) is odd will cancel out, and the terms where \(k\) is even will double:

\[ = 2 \left( \binom{5}{0} (\sqrt{2})^{5} + \binom{5}{2} (\sqrt{2})^{3} + \binom{5}{4} (\sqrt{2})^{1} \right) \]

### Step 4: Calculate the Binomial Coefficients

Now we calculate the binomial coefficients:

- \(\binom{5}{0} = 1\)
- \(\binom{5}{2} = 10\)
- \(\binom{5}{4} = 5\)

### Step 5: Substitute and Simplify

Substituting these values into the expression gives:

\[ = 2 \left( 1 \cdot (\sqrt{2})^{5} + 10 \cdot (\sqrt{2})^{3} + 5 \cdot (\sqrt{2})^{1} \right) \]

Calculating the powers of \(\sqrt{2}\):

- \((\sqrt{2})^{5} = 2^{5/2} = 4\sqrt{2}\)
- \((\sqrt{2})^{3} = 2^{3/2} = 2\sqrt{2}\)
- \((\sqrt{2})^{1} = \sqrt{2}\)

Now substituting these values:

\[ = 2 \left( 4\sqrt{2} + 10 \cdot 2\sqrt{2} + 5\sqrt{2} \right) \]
\[ = 2 \left( 4\sqrt{2} + 20\sqrt{2} + 5\sqrt{2} \right) \]
\[ = 2 \left( 29\sqrt{2} \right) \]
\[ = 58\sqrt{2} \]

### Final Answer

Thus, the value of \((\sqrt{2} + 1)^{5} + (\sqrt{2} - 1)^{5}\) is:

\[ \boxed{58\sqrt{2}} \]