JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Determinants

The roots of $x^2+x+1=0$ are the complex cube roots of unity $\omega$ and ${\omega }^2$, with $1+\omega +{\omega }^2=0$, so $\alpha +\beta =-1$. Adding all three rows into the first makes every first-row entry $y+1+\alpha +\beta =y$, so $y$ factors out: $\Delta =y\left|\begin{array}{ccc}1& 1& 1\\ \omega & y+{\omega }^2& 1\\ {\omega }^2& 1& y+\omega \end{array}\right|$. Expanding this and using $1+\omega +{\omega }^2=0$ with ${\omega }^3=1$, the remaining determinant collapses to $y^2$, so $\Delta =y^3$.