JEE Main 2019 (Online) 9th April Morning ShiftQuestion 1
Let $\alpha$ and $\beta$ be the roots of the equation $x^{2}+x+1=0$. Then for $y\neq0$ in $R$, $\begin{vmatrix}y+1&\alpha&\beta\\\alpha&y+\beta&1\\\beta&1&y+\alpha\end{vmatrix}$ is equal to
Determinant
Solution
With $\alpha+\beta=-1$, $\alpha\beta=1$, $\alpha^2=\beta$, $\beta^2=\alpha$, expanding the determinant, all lower terms cancel and it equals $y^3$.