Using binomial theorem, write the value of `(a+b)^(6) + (a-b)^(6)` and hence find the value of `( sqrt(3) + sqrt(2) )^(6) + ( sqrt(3) - sqrt(2) )^(6)`.

Answer

To solve the problem, we will use the Binomial Theorem to find the value of \((a+b)^6 + (a-b)^6\) and then substitute \(a = \sqrt{3}\) and \(b = \sqrt{2}\) to find the specific value.

### Step 1: Apply the Binomial Theorem
According to the Binomial Theorem, we can expand \((a+b)^6\) and \((a-b)^6\) as follows:

\[ (a+b)^6 = \sum_{k=0}^{6} \binom{6}{k} a^{6-k} b^k \]
\[ = \binom{6}{0} a^6 b^0 + \binom{6}{1} a^5 b^1 + \binom{6}{2} a^4 b^2 + \binom{6}{3} a^3 b^3 + \binom{6}{4} a^2 b^4 + \binom{6}{5} a^1 b^5 + \binom{6}{6} a^0 b^6 \]

Calculating the coefficients:

\[ = 1 \cdot a^6 + 6 \cdot a^5 b + 15 \cdot a^4 b^2 + 20 \cdot a^3 b^3 + 15 \cdot a^2 b^4 + 6 \cdot a b^5 + 1 \cdot b^6 \]

### Step 2: Expand \((a-b)^6\)
Now, we expand \((a-b)^6\):

\[ (a-b)^6 = \sum_{k=0}^{6} \binom{6}{k} a^{6-k} (-b)^k \]
\[ = \binom{6}{0} a^6 (-b)^0 + \binom{6}{1} a^5 (-b)^1 + \binom{6}{2} a^4 (-b)^2 + \binom{6}{3} a^3 (-b)^3 + \binom{6}{4} a^2 (-b)^4 + \binom{6}{5} a^1 (-b)^5 + \binom{6}{6} a^0 (-b)^6 \]

Calculating the coefficients:

\[ = 1 \cdot a^6 - 6 \cdot a^5 b + 15 \cdot a^4 b^2 - 20 \cdot a^3 b^3 + 15 \cdot a^2 b^4 - 6 \cdot a b^5 + 1 \cdot b^6 \]

### Step 3: Add the Two Expansions
Now we add \((a+b)^6\) and \((a-b)^6\):

\[ (a+b)^6 + (a-b)^6 = \left(1 \cdot a^6 + 6 \cdot a^5 b + 15 \cdot a^4 b^2 + 20 \cdot a^3 b^3 + 15 \cdot a^2 b^4 + 6 \cdot a b^5 + 1 \cdot b^6\right) + \left(1 \cdot a^6 - 6 \cdot a^5 b + 15 \cdot a^4 b^2 - 20 \cdot a^3 b^3 + 15 \cdot a^2 b^4 - 6 \cdot a b^5 + 1 \cdot b^6\right) \]

Combining like terms:

\[ = 2a^6 + 0 + 30a^4b^2 + 0 + 30a^2b^4 + 2b^6 \]
\[ = 2a^6 + 30a^4b^2 + 30a^2b^4 + 2b^6 \]

### Step 4: Substitute \(a = \sqrt{3}\) and \(b = \sqrt{2}\)
Now we substitute \(a = \sqrt{3}\) and \(b = \sqrt{2}\):

\[ = 2(\sqrt{3})^6 + 30(\sqrt{3})^4(\sqrt{2})^2 + 30(\sqrt{3})^2(\sqrt{2})^4 + 2(\sqrt{2})^6 \]

Calculating each term:

1. \((\sqrt{3})^6 = 27\)
2. \((\sqrt{3})^4 = 9\)
3. \((\sqrt{2})^2 = 2\)
4. \((\sqrt{3})^2 = 3\)
5. \((\sqrt{2})^4 = 4\)
6. \((\sqrt{2})^6 = 8\)

Now substituting these values:

\[ = 2(27) + 30(9)(2) + 30(3)(4) + 2(8) \]
\[ = 54 + 540 + 360 + 16 \]
\[ = 970 \]

### Final Answer
Thus, the value of \((\sqrt{3} + \sqrt{2})^6 + (\sqrt{3} - \sqrt{2})^6\) is \(\boxed{970}\).