Mathematics - Determinant Question with Solution | TestHub

It is given that the matrix $\left[\begin{array}{ccc}{\alpha }^2& \alpha & 1\\ 1& 1& 1\\ a& b& c\end{array}\right]$ is singular, so

$\left|\begin{array}{ccc}{\alpha }^2& \alpha & 1\\ 1& 1& 1\\ a& b& c\end{array}\right|=0$

$\Rightarrow {\alpha }^2\left(c-b\right)-\alpha \left(c-a\right)+\left(b-a\right)=0$

Above equation holds true if $\alpha =1$. Now, 

$\frac{{\left(a-c\right)}^2}{\left(b-a\right)\left(c-b\right)}+\frac{{\displaystyle {\left(b-a\right)}^2}}{\left(a-c\right){\displaystyle \left(c-b\right)}}+\frac{{\displaystyle {\left(c-b\right)}^2}}{\left(a-c\right){\displaystyle \left(b-a\right)}}$

$=\frac{{\left(a-b\right)}^3+{\left(b-c\right)}^3+{\left(c-a\right)}^3}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}$

$=\frac{3\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=3$

$\left[\because x^3+y^3+z^3=3xyz \mathrm{if} x+y+z=0\right]$