The sum of the co-efficient of all even degree terms in $$x$$ in the expansion of $$\left(x + \sqrt{x^{3} - 1}\right)^{6} + \left(x - \sqrt{x^{3} - 1}\right)^{6}$$, $$x \gt 1$$ is equal to:

We have to evaluate the expression

$$E(x)=\left(x+\sqrt{x^{3}-1}\right)^{6}+\left(x-\sqrt{x^{3}-1}\right)^{6},\qquad x\gt 1,$$

and then add the coefficients of all those terms in the expansion whose powers of $$x$$ are even.

First, recall the Binomial Theorem in the form $$\left(a+b\right)^{n}=\displaystyle\sum_{k=0}^{n}\binom{n}{k}a^{\,n-k}b^{\,k}.$$

Using this with $$a=x,\;b=\sqrt{x^{3}-1},\;n=6$$ gives

$$\left(x\pm\sqrt{x^{3}-1}\right)^{6}=\sum_{k=0}^{6}\binom{6}{k}\,x^{\,6-k}\left(\pm\sqrt{x^{3}-1}\right)^{k}.$$

Now we add the “+” and “−” versions. For odd $$k$$ the two terms cancel because one has a plus sign and the other a minus sign; for even $$k$$ they double because both signs become the same. Hence

$$E(x)=2\sum_{\substack{k=0\\k\text{ even}}}^{6}\binom{6}{k}\,x^{\,6-k}\Bigl(\sqrt{x^{3}-1}\Bigr)^{k}.$$

Put $$k=2m$$ (so $$m=0,1,2,3$$). We obtain

$$E(x)=2\sum_{m=0}^{3}\binom{6}{2m}\,x^{\,6-2m}\left(\sqrt{x^{3}-1}\right)^{2m}.$$

Because $$\left(\sqrt{x^{3}-1}\right)^{2m}=\left(x^{3}-1\right)^{m},$$ the expression becomes

$$E(x)=2\sum_{m=0}^{3}\binom{6}{2m}\,x^{\,6-2m}\left(x^{3}-1\right)^{m}.$$

Next, expand $$\left(x^{3}-1\right)^{m}$$ again by the Binomial Theorem:

$$\left(x^{3}-1\right)^{m}=\sum_{r=0}^{m}\binom{m}{r}(x^{3})^{r}(-1)^{\,m-r}=\sum_{r=0}^{m}\binom{m}{r}(-1)^{\,m-r}x^{\,3r}.$$

Substituting this inside $$E(x)$$ gives

$$E(x)=2\sum_{m=0}^{3}\binom{6}{2m}\sum_{r=0}^{m}\binom{m}{r}(-1)^{\,m-r}\,x^{\,6-2m+3r}.$$

Thus each term has coefficient

$$2\binom{6}{2m}\binom{m}{r}(-1)^{\,m-r}$$

and power of $$x$$ equal to $$n=6-2m+3r.$$ We now list all pairs $$(m,r)$$, compute $$n$$ and keep only those where $$n$$ is even.

• $$m=0$$: only $$r=0$$ is possible.
  Exponent $$n=6-0+0=6\;( \text{even}).$$
  Coefficient $$2\binom{6}{0}\binom{0}{0}(-1)^{0}=2.$$

• $$m=1$$: $$r=0,1.$$
  $$r=0$$: $$n=6-2+0=4\;(\text{even}),$$ coefficient $$2\binom{6}{2}\binom{1}{0}(-1)^{1}=2\cdot15\cdot1\cdot(-1)=-30.$$
  $$r=1$$: $$n=6-2+3=7\;(\text{odd}),$$ ignored.

• $$m=2$$: $$r=0,1,2.$$
  $$r=0$$: $$n=6-4+0=2\;(\text{even}),$$ coefficient $$2\binom{6}{4}\binom{2}{0}(-1)^{2}=2\cdot15\cdot1\cdot1=30.$$
  $$r=1$$: $$n=6-4+3=5\;(\text{odd}),$$ ignored.
  $$r=2$$: $$n=6-4+6=8\;(\text{even}),$$ coefficient $$2\binom{6}{4}\binom{2}{2}(-1)^{0}=2\cdot15\cdot1\cdot1=30.$$

• $$m=3$$: $$r=0,1,2,3.$$
  $$r=0$$: $$n=6-6+0=0\;(\text{even}),$$ coefficient $$2\binom{6}{6}\binom{3}{0}(-1)^{3}=2\cdot1\cdot1\cdot(-1)=-2.$$
  $$r=1$$: $$n=6-6+3=3\;(\text{odd}),$$ ignored.
  $$r=2$$: $$n=6-6+6=6\;(\text{even}),$$ coefficient $$2\binom{6}{6}\binom{3}{2}(-1)^{1}=2\cdot1\cdot3\cdot(-1)=-6.$$
  $$r=3$$: $$n=6-6+9=9\;(\text{odd}),$$ ignored.

We now add all coefficients corresponding to even exponents:

Constant term (degree 0): $$-2$$

Degree 2: $$+30$$

Degree 4: $$-30$$

Degree 6: $$2+(-6)=-4$$

Degree 8: $$+30$$

Sum of all these coefficients:

$$-2+30-30-4+30 = 24.$$

Hence, the required sum of coefficients of the even-degree terms is $$24$$.

Hence, the correct answer is Option C.