Let α be a root of the equation (a-c)x^2+(b-a)x+(c-b)=0 where a,b,c are distinct real numbers such that the matrix [ … ] is singular. Then the value…

It is given that the matrix $[\begin{matrix}{\alpha }^{2} & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c\end{matrix}]$[α21a​α1b​11c​] is singular, so

$$|\begin{matrix}{\alpha }^{2} & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c\end{matrix}|=0$$∣α21a​α1b​11c​∣=0

$$\Rightarrow {\alpha }^{2}(c-b)-\alpha (c-a)+(b-a)=0$$⇒α2(c−b)−α(c−a)+(b−a)=0

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

$$\frac{{(a-c)}^{2}}{(b-a)(c-b)}+\frac{{(b-a)}^{2}}{(a-c)(c-b)}+\frac{{(c-b)}^{2}}{(a-c)(b-a)}$$(b−a)(c−b)(a−c)2​+(a−c)(c−b)(b−a)2​+(a−c)(b−a)(c−b)2​

$$=\frac{{(a-b)}^{3}+{(b-c)}^{3}+{(c-a)}^{3}}{(a-b)(b-c)(c-a)}$$=(a−b)(b−c)(c−a)(a−b)3+(b−c)3+(c−a)3​

$$=\frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=3$$=(a−b)(b−c)(c−a)3(a−b)(b−c)(c−a)​=3

$$[\because {x}^{3}+{y}^{3}+{z}^{3}=3xyz\mathrm{if}x+y+z=0]$$[∵x3+y3+z3=3xyzifx+y+z=0]