In a triangle ABC, $a(b \cos C - c \cos B)$, is :

$a^2$ $b^2-c^2$ 0 None of these

$b^2-c^2$

$a =b \cos C + c \cos B$ $⇒a (b \cos C - c \cos B)=(b \cos C + c \cos B)×(b \cos C - c \cos B)$ $⇒b^2\cos^2C-c^2\cos^2B$ $=b^2(1- \sin^2C) - c^2(1- \sin^2 B)$ $=(b^2 - c^2)-b^2\sin^2C+c^2\sin^2B$ [by sine rule : $b \sin C = c \sin B$] $⇒b^2 - c^2$