JEE Advanced 2016 · Paper 1, Mathematics question 50

Split the third column as $1+x^3\cdot(1,8,27)$:
$$\Delta=x^3\begin{vmatrix}1&1&1\\2&4&1\\3&9&1\end{vmatrix}+x^6\begin{vmatrix}1&1&1\\2&4&8\\3&9&27\end{vmatrix}=2x^3+12x^6.$$

Solve. $12x^6+2x^3=10$. With $t=x^3$: $6t^2+t-5=0$, so $t=\frac56$ or $t=-1$. Each gives one real $x$.

Answer: 2