The total number of distinct $x \in R$ for which

$\begin{vmatrix} x & x^2 & 1+x^3\\2x& 4x^2 & 1+8x^3\\ 3x& 9x^2 & 1+27x^3 \end{vmatrix} = 10, $ is

2

The correct answer is option (4) : 2

$\begin{vmatrix} x & x^2 & 1+x^3\\2x& 4x^2 & 1+8x^3\\ 3x& 9x^2 & 1+27x^3 \end{vmatrix}=10$

$⇒x^3\begin{vmatrix} 1 & 1 & 1+x^3\\2 & 4 & 1+8x^3\\ 3 & 9 & 1+27x^3 \end{vmatrix}=10$

$⇒x^3\begin{vmatrix} 1 & 1 & 1\\2 & 4 & 1\\ 3 & 9 & 1\end{vmatrix}+x^3 \begin{vmatrix} 1 & 1 & x^3\\2 & 4 & 8x^3\\ 3 & 9 & 27x^3 \end{vmatrix}=10$

$⇒x^3\begin{vmatrix} 1 & 1 & 1\\2 & 4 & 1\\ 3 & 9 & 1\end{vmatrix}+x^6 \begin{vmatrix} 1 & 1 & 1\\2 & 4 & 8\\ 3 & 9 & 27 \end{vmatrix}=10$

Applying $R_2→R_2-R_1, R_3 → R_3-R_1 $ in first determinant and $C_2→C_2-C_1, C_3 → C_3 -C_1 $ in the second determinant,

we get

$⇒x^3\begin{vmatrix} 1 & 1 & 1\\1 & 3 & 0\\ 2 & 8 & 0\end{vmatrix}+x^6 \begin{vmatrix} 1 & 0 & 0\\2 & 2 & 6\\ 3 & 6 & 24 \end{vmatrix}=10$

$⇒(8-6)x^3+(48-36)x^6=10$

$⇒2x^3+12x^6=10$

$⇒6x^6+x^3-5=0$

$⇒(6x^3-5)(x^3+1)=0$

$⇒x^3=\frac{5}{6}$or, $x^3=-1 ⇒x=\left(\frac{5}{6}\right)^{1/3} $

or, $x=-1 $         $[∵x \in R]$