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# The sum of coefficient of all the even degree term in x in the expansion of ${{\text{(x + }}\sqrt {{{\text{x}}^{\text{3}}}{\text{ - 1}}} {\text{)}}^{\text{6}}}{\text{ + (x - }}\sqrt {{{\text{x}}^{\text{3}}}{\text{ - 1}}} {{\text{)}}^{\text{6}}}{\text{,x > 1}}$ is equal to $A.\;32 \\ B.\;26 \\ C.\;29 \\ D.\;24 \\$

Hint: Firstly we apply the binomial expansion formula as ${{\text{(x + y)}}^{\text{n}}}{\text{ = }}\sum\limits_{{\text{r = 0}}}^n {{}^{\text{n}}{{\text{C}}_{\text{r}}}{{\text{x}}^{{\text{n - r}}}}{{\text{y}}^{\text{r}}}}$and expand the above given expression. And we have to only consider the coefficient with even power of x and summation of all that will be our required answer.

As $n = 6,0 < r < 6$
First of all , expanding the above expression using the formula ${(x + y)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{x^{n - r}}{y^r}}$and on taking similar terms common we can obtained the expression as
${(x + \sqrt {{x^3} - 1} )^6} + {(x - \sqrt {{x^3} - 1} )^6} \\ = \sum\limits_{r = 0}^6 {{}^6{C_r}{x^{6 - r}}{{(\sqrt {{x^3} - 1} )}^r}} + \sum\limits_{r = 0}^6 {{}^6{C_r}{x^{6 - r}}{{(\sqrt {{x^3} - 1} )}^r}{{( - 1)}^r}} \\ = \sum\limits_{r = 0}^6 {{}^6{C_r}{x^{6 - r}}[{{(\sqrt {{x^3} - 1} )}^r} + {{( - 1)}^r}{{(\sqrt {{x^3} - 1} )}^r}]} \\$
From the above expression we can see that for all the odd values of r complete expression will be 0.
Now, solving by taking only even values of r, expression can be simplified to
$= 2[\sum\limits_{r = 2n,n = 0}^{n = 3} {{}^6{C_r}{x^{6 - r}}{{(\sqrt {{x^3} - 1} )}^r}} ]$
Now putting the even values of r, as we know ${\text{n = 6,0 < r < 6}}$
$= 2[{}^6{C_0}{x^{6 - 0}}{(\sqrt {{x^3} - 1} )^0} + {}^6{C_2}{x^{6 - 2}}{(\sqrt {{x^3} - 1} )^2} + {}^6{C_4}{x^{6 - 4}}{(\sqrt {{x^3} - 1} )^4} + {}^6{C_6}{x^0}{(\sqrt {{x^3} - 1} )^6} \\ = 2[{}^6{C_0}{x^6} + {}^6{C_2}{x^4}({x^3} - 1) + {}^6{C_4}{x^2}{({x^3} - 1)^2} + {}^6{C_6}{({x^3} - 1)^3} \\$
Now, directly writing the value of ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ and expanding various algebraic expressions in order to select the coefficients of the term with even power of x only.
$= 2[1.{x^6} + 15.({x^7} - {x^4}) + 15.({x^6} - 2{x^3} + 1) + 1.{({x^3} - 1)^3}]$
Now, also calculation the expanded value of
${({x^3} - 1)^3} = {x^9} + 3{({x^3})^2}( - 1) + 3{x^3}{( - 1)^2} + 1 \\ = {x^9} - 3{x^6} + 3{x^3} + 1 \\$
Now substituting the obtained expanded algebraic value in main binomial expansion expression,
$= 2[1.{x^6} + 15.({x^7} - {x^4}) + 15.({x^8} - 2{x^5} + {x^2}) + 1.({x^9} - 3{x^6} + 3{x^3} + 1)] \\ = 2[{x^6} + 15{x^7} - 15{x^4} + 15{x^8} - 30{x^5} + 15{x^2} + {x^9} - 3{x^6} + 3{x^3} + 1] \\$
Now, add all the coefficient values of even power of x.
$= 2(1 - 15 + 15 + 15 - 1-3) \\ = 2(15 - 3) \\ = 2(12) \\ = 24 \\$

Hence , option (d) is our required answer.

Note: Binomial theorem, statement that for any positive integer n, the nth power of the sum of two numbers a and b may be expressed as the sum of n + 1 terms of the form.
The coefficients of the terms in the expansion are the binomial coefficients (kn). The theorem and its generalizations can be used to prove results and solve problems in combinatorics, algebra, calculus, and many other areas of mathematics.