The value of `(sqrt5+1)^(5)-(sqrt5-1)^(5)` is

Options

A. 252
B. 352
C. 452
D. 552

Answer

To solve the expression \((\sqrt{5}+1)^{5} - (\sqrt{5}-1)^{5}\), we will use the Binomial Theorem to expand both terms and then simplify.

### Step-by-Step Solution:

1. **Apply the Binomial Theorem**:
The Binomial Theorem states that:
\[ (x + a)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} a^k \]
We will apply this to both \((\sqrt{5}+1)^{5}\) and \((\sqrt{5}-1)^{5}\).

2. **Expand \((\sqrt{5}+1)^{5}\)**:
Using the Binomial Theorem:
\[ (\sqrt{5}+1)^{5} = \sum_{k=0}^{5} \binom{5}{k} (\sqrt{5})^{5-k} (1)^{k} \]
This expands to:
\[ = \binom{5}{0} (\sqrt{5})^{5} + \binom{5}{1} (\sqrt{5})^{4} \cdot 1 + \binom{5}{2} (\sqrt{5})^{3} \cdot 1^2 + \binom{5}{3} (\sqrt{5})^{2} \cdot 1^3 + \binom{5}{4} (\sqrt{5})^{1} \cdot 1^4 + \binom{5}{5} (1)^{5} \]
Simplifying this gives:
\[ = 5\sqrt{5}^5 + 10\sqrt{5}^4 + 10\sqrt{5}^3 + 5\sqrt{5}^2 + 1 \]

3. **Expand \((\sqrt{5}-1)^{5}\)**:
Similarly, we expand:
\[ (\sqrt{5}-1)^{5} = \sum_{k=0}^{5} \binom{5}{k} (\sqrt{5})^{5-k} (-1)^{k} \]
This expands to:
\[ = \binom{5}{0} (\sqrt{5})^{5} - \binom{5}{1} (\sqrt{5})^{4} + \binom{5}{2} (\sqrt{5})^{3} - \binom{5}{3} (\sqrt{5})^{2} + \binom{5}{4} (\sqrt{5})^{1} - \binom{5}{5} (1)^{5} \]
Simplifying this gives:
\[ = 5\sqrt{5}^5 - 10\sqrt{5}^4 + 10\sqrt{5}^3 - 5\sqrt{5}^2 + 1 \]

4. **Combine the Two Expansions**:
Now, we subtract the two expansions:
\[ (\sqrt{5}+1)^{5} - (\sqrt{5}-1)^{5} = (5\sqrt{5}^5 + 10\sqrt{5}^4 + 10\sqrt{5}^3 + 5\sqrt{5}^2 + 1) - (5\sqrt{5}^5 - 10\sqrt{5}^4 + 10\sqrt{5}^3 - 5\sqrt{5}^2 + 1) \]
Simplifying this gives:
\[ = 20\sqrt{5}^4 + 10\sqrt{5}^2 \]

5. **Final Simplification**:
We can factor out \(10\):
\[ = 10(2\sqrt{5}^4 + \sqrt{5}^2) \]
Since \(\sqrt{5}^2 = 5\), we have:
\[ = 10(2 \cdot 25 + 5) = 10(50 + 5) = 10 \cdot 55 = 550 \]

### Final Answer:
The value of \((\sqrt{5}+1)^{5} - (\sqrt{5}-1)^{5}\) is \(550\).