Question

The sum of coefficients of even powers of $x$ in ${\left(x+\sqrt{x^3-1}\right)}^6+{\left(x-\sqrt{x^3-1}\right)}^6$

• A. $23$
• B. $24$
• C. $18$
• D. $21$

Answer 1

The correct option is B

$24$

Explanation for correct option:

We know that ${\left(a+b\right)}^6+{\left(a-b\right)}^6=2\left(a^6+15a^4{b}^2+15a^2{b}^4+b^6\right)$

Expanding the help of binomial theorem we get

${\left(x+\sqrt{x^3-1}\right)}^6+{\left(x-\sqrt{x^3-1}\right)}^6$

$$\begin{gathered} =2\left[x^6+15x^4\left(x^3-1\right)+15x^2{\left(x^3-1\right)}^2+{\left(x^3-1\right)}^3\right] \\ =2\left[x^6+15x^7-15x^4+15x^8-30x^5+15x^2+x^9-3x^6+3x^3-1\right]\left[\because {\left(a-b\right)}^3=a^3-3a^{2}b+3a{b}^2+b^3\right] \end{gathered}$$

Term with even power is

$=2\left[x^6-15x^4+15x^8+15x^2-3x^6-1\right]$

Therefore the sum of the coefficient is

$$\begin{gathered} =2\left(1-15+15+15-3-1\right) \\ =24 \end{gathered}$$

Hence option (B) is correct.