JEE Main 2019 — Mathematics Algebra

Roots of the equation ${x}^{2}+x+1=0$x2+x+1=0 are $\alpha$α and $\beta$β and we know that the sum of roots of a quadratic equation $a{x}^{2}+bx+c=0$ax2+bx+c=0 is $-\frac{b}{a}$−ab​ and $\frac{c}{a}$ac​ respectively.

$\Rightarrow \alpha +\beta =-1$⇒α+β=−1 and $\alpha \beta =1.$αβ=1.

Now, let $\Delta =|\begin{matrix}y+1 & \alpha & \beta \\ \alpha & y+\beta & 1 \\ \beta & 1 & y+\alpha \end{matrix}|$Δ=∣y+1αβ​αy+β1​β1y+α​∣

Applying ${R}_{1}\rightarrow {R}_{1}+{R}_{2}+{R}_{3},$R1​→R1​+R2​+R3​, we get

$$\Delta =|\begin{matrix}y+1+\alpha +\beta & y+1+\alpha +\beta & y+1+\alpha +\beta \\ \alpha & y+\beta & 1 \\ \beta & 1 & y+\alpha \end{matrix}|$$Δ=∣y+1+α+βαβ​y+1+α+βy+β1​y+1+α+β1y+α​∣

$$=(y+1+\alpha +\beta )|\begin{matrix}1 & 1 & 1 \\ \alpha & y+\beta & 1 \\ \beta & 1 & y+\alpha \end{matrix}|$$=(y+1+α+β)∣1αβ​1y+β1​11y+α​∣

$$=(y+1+(-1))[1{(y+\beta )(y+\alpha )-1}-1{\alpha (y+\alpha )-\beta }+1{\alpha -\beta (y+\beta )}]$$=(y+1+(−1))[1(y+β)(y+α)−1−1α(y+α)−β+1α−β(y+β)]

$$=y[{y}^{2}+(\alpha +\beta )y+\alpha \beta -y\alpha -{\alpha }^{2}+\beta +\alpha -\beta y-{\beta }^{2}]$$=y[y2+(α+β)y+αβ−yα−α2+β+α−βy−β2]

$$=y[{y}^{2}+(\alpha +\beta )y+\alpha \beta -y(\alpha +\beta +1)-({\alpha }^{2}+{\beta }^{2})+(\alpha +\beta )]$$=y[y2+(α+β)y+αβ−y(α+β+1)−(α2+β2)+(α+β)]

On putting the values of \alpha +\beta &\alpha \beta , we get

$$=y[{y}^{2}+(-1)y+1-y(-1+1)-({(\alpha +\beta )}^{2}-2\alpha \beta )+(-1)]$$=y[y2+(−1)y+1−y(−1+1)−((α+β)2−2αβ)+(−1)]

$$=y[{y}^{2}-y+1-({(-1)}^{2}-2)-1]$$=y[y2−y+1−((−1)2−2)−1]

$$=y[{y}^{2}-y+1]$$=y[y2−y+1]

$={y}^{3}$=y3 (on simplifying)