Evalute: ` (sqrt(3)+1)^(5) -(sqrt(3)-1)^(5)`

Answer

To evaluate the expression \((\sqrt{3}+1)^{5} - (\sqrt{3}-1)^{5}\), we can use the Binomial Theorem.

### Step-by-Step Solution:

1. **Apply the Binomial Theorem**:
The Binomial Theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
For our expression, we will expand both \((\sqrt{3}+1)^{5}\) and \((\sqrt{3}-1)^{5}\).

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

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

4. **Combine the Results**:
Now we can subtract the two expansions:
\[ (\sqrt{3}+1)^{5} - (\sqrt{3}-1)^{5} \]
Notice that all even-powered terms will cancel out, and we will be left with:
\[ = 2 \left( 5 \cdot 3^{2} + 10 \cdot 3^{3/2} + 5 \cdot \sqrt{3} + 1 \right) \]

5. **Calculate the Result**:
Now we calculate the coefficients:
\[ = 2 \left( 5 \cdot 9 + 10 \cdot 3\sqrt{3} + 5\sqrt{3} + 1 \right) \]
\[ = 2 \left( 45 + 10\sqrt{3} + 5\sqrt{3} + 1 \right) \]
\[ = 2 \left( 46 + 15\sqrt{3} \right) \]
\[ = 92 + 30\sqrt{3} \]

### Final Answer:
\[ (\sqrt{3}+1)^{5} - (\sqrt{3}-1)^{5} = 92 + 30\sqrt{3} \]