If $$\alpha$$ and $$\beta$$, be the coefficients of $$x^4$$ and $$x^2$$, respectively in the expansion of $$\left(x + \sqrt{x^2 - 1}\right)^6 + \left(x - \sqrt{x^2 - 1}\right)^6$$, then

To solve for the coefficients $$\alpha$$ and $$\beta$$, we can use the binomial identity for the sum of two powers:

$$(a+b)^n + (a-b)^n = 2 \left[ \binom{n}{0}a^n + \binom{n}{2}a^{n-2}b^2 + \binom{n}{4}a^{n-4}b^4 + \dots \right]$$

1. Simplification of the Expression

Given the expression $$\left(x + \sqrt{x^2 - 1}\right)^6 + \left(x - \sqrt{x^2 - 1}\right)^6$$, let a = x and $$b=\sqrt{\ x^2-1}$$. Applying the identity with $$n=6$$:

$$E = 2 \left[ \binom{6}{0}x^6 + \binom{6}{2}x^4(x^2-1)^1 + \binom{6}{4}x^2(x^2-1)^2 + \binom{6}{6}(x^2-1)^3 \right]$$

$$E = 2 \left[ x^6 + 15x^4(x^2-1) + 15x^2(x^2-1)^2 + (x^2-1)^3 \right]$$

2. Expanding the Terms

Now, expand the powers of $$\left(x^2-1\right)$$:

• $$15x^4\left(x^2-1\right)\ =\ 15x^6\ -\ 15x^4$$
  
• $$15x^4\left(x^2-1\right)^2\ =\ 15x^6\ -\ 30x^4+15x^2$$
  
• $$\left(x^2-1\right)^3\ =\ x^6-3x^4+3x^2-1$$
  

Substitute these back into the expression:

$$E = 2 \left[ (x^6) + (15x^6 - 15x^4) + (15x^6 - 30x^4 + 15x^2) + (x^6 - 3x^4 + 3x^2 - 1) \right]$$

3. Combining Like Terms

Group the terms by powers of x:

• $$x^6$$ terms: $$1+15+15+1\ =\ 32$$
• $$x^4$$ terms: $$-15-30-3\ =\ -48$$
• $$x^2$$ terms: $$15+3\ =\ 18$$
• Constant: -1

$$E\ =\ 2\left[32x^6\ -\ 48x^4+18x^2-1\right]$$

4. Finding $$\alpha$$ and $$\beta$$

From the expanded form, we identify the coefficients:

• $$\alpha$$  =  -96
• $$\beta$$  =  36

Now, calculate $$\alpha - \beta$$:

$$\alpha - \beta = -96 - 36 = -132$$

Final Answer:

The correct option is D, $$\alpha - \beta = -132$$.