The total number of distinct $x∈ R$ 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

We have,

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

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

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

$⇒x^3\begin{vmatrix}1 &1 &1\\2 &4 &1\\3& 9& 1\end{vmatrix}+x^6\begin{vmatrix}1 &1 &1\\2 &4 &8\\3& 9& 24\end{vmatrix}=10$

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

$⇒(8-6) x^3+(48-36) x^6 =10$

$⇒2x^3+12x^6 = 10$

$⇒6x^6+x^3-5=0$

$⇒(6x^3-5)(x^3+1)=0$

$⇒x^3=\frac{5}{6}$ or, $x^3=-1⇒x=(\frac{5}{6})^{1/3}$ or, $x=-1$  $[∵x∈ R]$