# Solve this following

Question:

Let $\alpha$ and $\beta$ be the roots of the equation $x^{2}+x+1=0$. Then for $y \neq 0$ in $R$,

$\left|\begin{array}{ccc}y+1 & \alpha & \beta \\ \alpha & y+\beta & 1 \\ \beta & 1 & y+\alpha\end{array}\right|$ is equal to

1. $y^{3}$

2. $y^{3}-1$

3. $y\left(y^{2}-1\right)$

4. $y\left(y^{2}-3\right)$

Correct Option: 1

Solution:

Roots of the equation $x^{2}+x+1=0$ are $\alpha=$ $\omega$ and $\beta=\omega^{2}$

where $\omega, \omega^{2}$ are complex cube roots of unity

$\therefore \Delta=\left|\begin{array}{ccc}y+1 & \omega & \omega^{2} \\ \omega & y+\omega^{2} & 1 \\ \omega^{2} & 1 & y+\omega\end{array}\right|$

$\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}$

$\Rightarrow \Delta=\mathrm{y}\left|\begin{array}{ccc}1 & 1 & 1 \\ \omega & \mathrm{y}+\omega^{2} & 1 \\ \omega^{2} & 1 & \mathrm{y}+\omega\end{array}\right|$

Expanding along $R_{1}$, we get

$\Delta=\mathrm{y} \cdot \mathrm{y}^{2} \Rightarrow \mathrm{D}=\mathrm{y}^{3}$