In any ΔABC, prove that

Left hand side,

$a(b \cos C-c \cos B)$

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

$=a b \frac{a^{2}+b^{2}-c^{2}}{2 a b}-a c \frac{a^{2}+c^{2}-b^{2}}{2 a c}\left[A s, \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \& \cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}\right]$

$=\frac{a^{2}+b^{2}-c^{2}}{2}-\frac{a^{2}+c^{2}-b^{2}}{2}$

$=\frac{a^{2}+b^{2}-c^{2}-a^{2}-c^{2}+b^{2}}{2}$

$=\frac{2\left(b^{2}-c^{2}\right)}{2}$

$=b^{2}-c^{2}$

$=$ Right hand side. [Proved]