Mathematics - Determinant Question with Solution | TestHub

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 $-\frac{b}{a}$ and $\frac{c}{a}$ respectively.

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

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

Applying $R_1\to R_1+R_2+R_3,$ we get

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

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

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

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

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

On putting the values of $\alpha +\beta \& \alpha \beta ,$ 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)