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

$a =b \cos C + c \cos B$

$⇒a (b \cos C - c \cos B)=(b \cos C + c \cos B)×(b \cos C - c \cos B)$

$⇒b^2\cos^2C-c^2\cos^2B$

$=b^2(1- \sin^2C) - c^2(1- \sin^2 B)$

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

$⇒b^2 - c^2$