Let α be a root of the equation (a - c) x ^ 2 + (b - a)x + (c - b)...

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

\left|$\begin{cases}{\alpha}^{2} & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c\end{cases}$\right| = 0$</p><p>$⇒ {\alpha}^{2}\left(c - b\right) - \alpha\left(c - a\right) + \left(b - a\right) = 0$</p><p>Above equation holds true if$\alpha = 1$. Now, </p><p>$\dfrac{{\left(a - c\right)}^{2}}{\left(b - a\right)\left(c - b\right)} + \dfrac{{\left(b - a\right)}^{2}}{\left(a - c\right)\left(c - b\right)} + \dfrac{{\left(c - b\right)}^{2}}{\left(a - c\right)\left(b - a\right)}$</p><p>$= \dfrac{{\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)}$</p><p>$= \dfrac{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$</p><p>\left[∵ {x}^{3} + {y}^{3} + {z}^{3} = 3xyzifx + y + z = 0\right]$