The sum of the co-efficient of all odd degree terms in the expansion of $$\left(x + \sqrt{x^3 - 1}\right)^5 + \left(x - \sqrt{x^3 - 1}\right)^5$$, $$(x > 1)$$ is:

We have to find the sum of the coefficients of all those terms whose powers of $$x$$ are odd in the expression

$$\left(x+\sqrt{x^{3}-1}\right)^{5}+\left(x-\sqrt{x^{3}-1}\right)^{5},\qquad x>1.$$

Put $$y=\sqrt{x^{3}-1}$$ just for convenience. The given expression becomes

$$E=(x+y)^{5}+(x-y)^{5}.$$

First recall the binomial formula:

$$(a+b)^{5}=a^{5}+5a^{4}b+10a^{3}b^{2}+10a^{2}b^{3}+5ab^{4}+b^{5}.$$

Using it for $$(x+y)^{5}$$ we get

$$x^{5}+5x^{4}y+10x^{3}y^{2}+10x^{2}y^{3}+5xy^{4}+y^{5}.$$

For $$(x-y)^{5}$$ every odd power of $$y$$ picks up a minus sign, so we obtain

$$x^{5}-5x^{4}y+10x^{3}y^{2}-10x^{2}y^{3}+5xy^{4}-y^{5}.$$

Now add the two expansions term-by-term:

$$E=\bigl[x^{5}+x^{5}\bigr]+\bigl[5x^{4}y-5x^{4}y\bigr]+\bigl[10x^{3}y^{2}+10x^{3}y^{2}\bigr]+\bigl[10x^{2}y^{3}-10x^{2}y^{3}\bigr]+\bigl[5xy^{4}+5xy^{4}\bigr]+\bigl[y^{5}-y^{5}\bigr].$$

All parts containing odd powers of $$y$$ cancel, while the even ones double. Hence

$$E=2x^{5}+20x^{3}y^{2}+10xy^{4}.$$

Next rewrite everything solely in powers of $$x$$. Because $$y=\sqrt{x^{3}-1}$$, we have $$y^{2}=x^{3}-1$$ and therefore $$y^{4}=(y^{2})^{2}=(x^{3}-1)^{2}=x^{6}-2x^{3}+1.$$ Substitute these:

$$E=2x^{5}+20x^{3}(x^{3}-1)+10x\bigl(x^{6}-2x^{3}+1\bigr).$$

Work out each product fully:

$$20x^{3}(x^{3}-1)=20x^{6}-20x^{3},$$ $$10x\bigl(x^{6}-2x^{3}+1\bigr)=10x^{7}-20x^{4}+10x.$$

Collect every term:

$$E=10x^{7}+20x^{6}+2x^{5}-20x^{4}-20x^{3}+10x.$$

The odd powers of $$x$$ present here are $$x^{7},x^{5},x^{3}$$ and $$x^{1}$$. Their respective coefficients are

$$10,\;2,\;-20,\;10.$$

Add these coefficients:

$$10+2-20+10=2.$$

So the sum of the coefficients of all odd-degree terms is $$2$$.

Hence, the correct answer is Option A.