Question

Let $\alpha$ 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 $\begin{bmatrix}\alpha^2 & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c\end{bmatrix}$ is singular. Then, the value of $\frac{(a-c)^2}{(b-a)(c-b)}+\frac{(b-a)^2}{(a-c)(c-b)}+\frac{(c-b)^2}{(a-c)(b-a)}$ is

• A. 6
• B. 3
• C. 9
• D. 12

Answer 1

Answer: (B)

$\Delta=0=\begin{vmatrix}\alpha^2 & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c\end{vmatrix}$
$\Rightarrow \alpha^2(c-b)-\alpha(c-a)+(b-a)=0$
It is singular when $\alpha=1$
$ \frac{(a-c)^2}{(b-a)(c-b)}+\frac{(b-a)^2}{(a-c)(c-b)}+\frac{(c-b)^2}{(a-c)(b-a)}$
$ \frac{(a-b)^3+(b-c)^3+(c-a)^3}{(a-b)(b-c)(c-a)} $
$ =3 \frac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=3$