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)

3

Answer 2

The correct answer is (B) : 3
Δ=0=||⇒a21a?a1b?11c?||?
⇒a2(c-b)-a(c-a)+(b-a)=0
It is singular when a=1
(b-a)(c-b)(a-c)2?+(a-c)(c-b)(b-a)2?+(a-c)(b-a)(c-b)2?
(a-b)(b-c)(c-a)(a-b)3+(b-c)3+(c-a)3?
=3(a-b)(b-c)(c-a)(a-b)(b-c)(c-a)?=3