Let α and β be the roots of the equation x ^ 2 + x + 1 = 0 . Then...

Roots of the equation ${x}^{2} + x + 1 = 0$ are $\alpha$ and $\beta$ and we know that the sum of roots of a quadratic equation $a{x}^{2} + bx + c = 0$ is $- \dfrac{b}{a}$ and $\dfrac{c}{a}$ respectively.

$⇒ \alpha + \beta = - 1$ and $\alpha\beta = 1 .$

Now, let ∆ = \left|\begin{cases}y + 1 & \alpha & \beta \\ \alpha & y + \beta & 1 \\ \beta & 1 & y + \alpha\end{cases}\right|

Applying ${R}_{1} \to {R}_{1} + {R}_{2} + {R}_{3},$ we get

∆ = \left|\begin{cases}y + 1 + \alpha + \beta & y + 1 + \alpha + \beta & y + 1 + \alpha + \beta \\ \alpha & y + \beta & 1 \\ \beta & 1 & y + \alpha\end{cases}\right|

= \left(y + 1 + \alpha + \beta\right)\left|\begin{cases}1 & 1 & 1 \\ \alpha & y + \beta & 1 \\ \beta & 1 & y + \alpha\end{cases}\right|

$= \left(y + 1 + \left(- 1\right)\right)\left[1\left\{\left(y + \beta\right)\left(y + \alpha\right) - 1\right\} - 1\left\{\alpha\left(y + \alpha\right) - \beta\right\} + 1\left\{\alpha - \beta\left(y + \beta\right)\right\}\right]$

$= y\left[{y}^{2} + \left(\alpha + \beta\right)y + \alpha\beta - y\alpha - {\alpha}^{2} + \beta + \alpha - \betay - {\beta}^{2}\right]$

$= y\left[{y}^{2} + \left(\alpha + \beta\right)y + \alpha\beta - y\left(\alpha + \beta + 1\right) - \left({\alpha}^{2} + {\beta}^{2}\right) + \left(\alpha + \beta\right)\right]$

On putting the values of $\alpha + \beta \text{ and } \alpha\beta,$ we get

$= y\left[{y}^{2} + \left(- 1\right)y + 1 - y\left(- 1 + 1\right) - \left({\left(\alpha + \beta\right)}^{2} - 2\alpha\beta\right) + \left(- 1\right)\right]$

$= y\left[{y}^{2} - y + 1 - \left({\left(- 1\right)}^{2} - 2\right) - 1\right]$

$= y\left[{y}^{2} - y + 1\right]$

$= {y}^{3}$ (on simplifying)