Answer:
Option A
Explanation:
Given, $\begin{bmatrix}x & x^{2} &1+x^{3} \\2x & 4x^{2}& 1+8x^{3} \\3x&9x^{2}&1+27x^{3} \end{bmatrix}=10 $
= $\Rightarrow x.x^{2}\begin{bmatrix}1 & 1 &1+x^{3} \\2 & 4& 1+8x^{3} \\3&9&1+27x^{3} \end{bmatrix}=10 $
Apply $R_{2}\rightarrow R_{2}-2R_{1}$ and $R_{3}\rightarrow R_{3}-3R_{1}$
we get,
$x^{3}\begin{bmatrix}1 & 1 &1+x^{3} \\0 & 2& -1+6x^{3} \\0&6&-2+24x^{3} \end{bmatrix}=10$
$\Rightarrow$ $x^{3}\begin{bmatrix}2 & 6x^{3} &-1\\6 & 24x^{3}&-2 \end{bmatrix}=10$
$\Rightarrow$ $x^{3} (48x^{3}-4-36x^{3}+6)=10$
$\Rightarrow$ $12x^{6}+2x^{3}=10$
$\Rightarrow$ $6x^{6}+x^{3}-5=0$
$\Rightarrow$ $x^{3}=\frac{5}{6},-1$
$\therefore$ $x=\left(\frac{5}{6}\right)^{1/3},-1$
Hence, the number of real solutions is 2
