Evaluate using binomial theorem:
`(i) (sqrt(2)+1)^(6) +(sqrt(2)-1)^(6)`
`(ii) (sqrt(5)+sqrt(2))^(4)-(sqrt(5)-sqrt(2))^(4)`

Explanation

To solve the given problems using the Binomial Theorem, we will follow a systematic approach for each part.

### Part (i): Evaluate \((\sqrt{2}+1)^{6} + (\sqrt{2}-1)^{6}\)

1. **Apply the Binomial Theorem**:
The Binomial Theorem states that:
\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]
For \((\sqrt{2}+1)^{6}\), we have \(a = \sqrt{2}\), \(b = 1\), and \(n = 6\).

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

This gives us:
\[ = \binom{6}{0} (\sqrt{2})^{6} + \binom{6}{1} (\sqrt{2})^{5} + \binom{6}{2} (\sqrt{2})^{4} + \binom{6}{3} (\sqrt{2})^{3} + \binom{6}{4} (\sqrt{2})^{2} + \binom{6}{5} (\sqrt{2})^{1} + \binom{6}{6} (1)^{6} \]

Simplifying:
\[ = \binom{6}{0} \cdot 2^{3} + \binom{6}{1} \cdot 2^{5/2} + \binom{6}{2} \cdot 2^{4} + \binom{6}{3} \cdot 2^{3/2} + \binom{6}{4} \cdot 2 + \binom{6}{5} \cdot \sqrt{2} + \binom{6}{6} \]

2. **Calculate \((\sqrt{2}-1)^{6}\)**:
Similarly, for \((\sqrt{2}-1)^{6}\):
\[ (\sqrt{2}-1)^{6} = \sum_{r=0}^{6} \binom{6}{r} (\sqrt{2})^{6-r} (-1)^r \]

This gives us:
\[ = \binom{6}{0} (\sqrt{2})^{6} - \binom{6}{1} (\sqrt{2})^{5} + \binom{6}{2} (\sqrt{2})^{4} - \binom{6}{3} (\sqrt{2})^{3} + \binom{6}{4} (\sqrt{2})^{2} - \binom{6}{5} (\sqrt{2}) + \binom{6}{6} \]

3. **Combine the two expansions**:
Now, we add the two expansions:
\[ (\sqrt{2}+1)^{6} + (\sqrt{2}-1)^{6} \]

Notice that terms with odd powers of \(\sqrt{2}\) will cancel out:
\[ = 2\left(\binom{6}{0} \cdot 2^{3} + \binom{6}{2} \cdot 2^{4} + \binom{6}{4} \cdot 2\right) + 2 \]

4. **Calculate the coefficients**:
\[ = 2\left(1 \cdot 8 + 15 \cdot 4 + 15 \cdot 2\right) + 2 \]
\[ = 2(8 + 60 + 30) + 2 = 2(98) + 2 = 196 + 2 = 198 \]

### Final answer for Part (i):
\[ (\sqrt{2}+1)^{6} + (\sqrt{2}-1)^{6} = 198 \]

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### Part (ii): Evaluate \((\sqrt{5}+\sqrt{2})^{4} - (\sqrt{5}-\sqrt{2})^{4}\)

1. **Apply the Binomial Theorem**:
For \((\sqrt{5}+\sqrt{2})^{4}\):
\[ (\sqrt{5}+\sqrt{2})^{4} = \sum_{r=0}^{4} \binom{4}{r} (\sqrt{5})^{4-r} (\sqrt{2})^{r} \]

This gives us:
\[ = \binom{4}{0} (\sqrt{5})^{4} + \binom{4}{1} (\sqrt{5})^{3} (\sqrt{2}) + \binom{4}{2} (\sqrt{5})^{2} (\sqrt{2})^{2} + \binom{4}{3} (\sqrt{5}) (\sqrt{2})^{3} + \binom{4}{4} (\sqrt{2})^{4} \]

2. **Calculate \((\sqrt{5}-\sqrt{2})^{4}\)**:
Similarly, for \((\sqrt{5}-\sqrt{2})^{4}\):
\[ (\sqrt{5}-\sqrt{2})^{4} = \sum_{r=0}^{4} \binom{4}{r} (\sqrt{5})^{4-r} (-\sqrt{2})^{r} \]

3. **Combine the two expansions**:
Now, we subtract the two expansions:
\[ (\sqrt{5}+\sqrt{2})^{4} - (\sqrt{5}-\sqrt{2})^{4} \]

Notice that terms with even powers of \(\sqrt{2}\) will cancel out:
\[ = 2\left(\binom{4}{1} (\sqrt{5})^{3} (\sqrt{2}) + \binom{4}{3} (\sqrt{5}) (\sqrt{2})^{3}\right) \]

4. **Calculate the coefficients**:
\[ = 2\left(4 (\sqrt{5})^{3} (\sqrt{2}) + 4 (\sqrt{5}) (\sqrt{2})^{3}\right) \]
\[ = 8\left(\sqrt{5}^{3} \sqrt{2} + \sqrt{5} \sqrt{2}^{3}\right) = 8\left(5\sqrt{5}\sqrt{2} + 2\sqrt{5}\sqrt{2}\right) = 8 \cdot 7\sqrt{10} = 56\sqrt{10} \]

### Final answer for Part (ii):
\[ (\sqrt{5}+\sqrt{2})^{4} - (\sqrt{5}-\sqrt{2})^{4} = 56\sqrt{10} \]

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