In ∆ABC, prove the following:

Solution:

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k \quad \ldots(1)$

$\mathrm{LHS}=\sin ^{3} A \cos (B-C)+\sin ^{3} B \cos (C-A)+\sin ^{3} C \cos (A-B)$

$=\sin ^{2} A\{\sin A \cos (B-C)\}+\sin ^{2} B\{\sin B \cos (C-A)\}+\sin ^{2} A\{\sin A \cos (A-B)\}$

$=\frac{a^{2}}{k^{2}}\{\sin A \cos (B-C)\}+\frac{b^{2}}{k^{2}}\{\sin B \cos (C-A)\}+\frac{c^{2}}{k^{2}}\{\sin A \cos (A-B)\} \quad$ [from (1)]

$=\frac{1}{2 k^{2}}\left[a^{2}\{2 \sin A \cos (B-C)\}+b^{2}\{2 \sin B \cos (C-A)\}+c^{2}\{2 \sin A \cos (A-B)\}\right]$

$=\frac{1}{2 k^{2}}\left[a^{2}\{2 \sin (\pi-(B+C)) \cos (B-C)\}+b^{2}\{2 \sin (\pi-(A+C)) \cos (C-A)\}+c^{2}\{2 \sin (\pi-(B+C)) \cos (A-B)\}\right]$

$=\frac{1}{2 k^{2}}\left[a^{2}\{2 \sin (B+C) \cos (B-C)\}+b^{2}\{2 \sin (C+A) \cos (C-A)\}+c^{2}\{2 \sin (A+B) \cos (A-B)\}\right]$

$=\frac{1}{2 k^{2}}\left[a^{2}\{\sin 2 B+\sin 2 C\}+b^{2}\{\sin 2 C+\sin 2 A\}+c^{2}\{\sin 2 A+\sin 2 B\}\right]$

$=\frac{1}{2 k^{2}}\left[2 a^{2}\{\sin B \cos B+\sin C \cos C\}+2 b^{2}\{\sin C \cos C+\sin A \cos A\}+2 c^{2}\{\sin A \cos A+\sin B \cos B\}\right]$

$=\frac{1}{\alpha^{3}}\left[2 a^{2}\{k \sin B \cos B+k \sin C \cos C\}+2 b^{2}\{k \sin C \cos C+k \sin A \cos A\}+2 c^{2}\{k \sin A \cos A+k \sin B \cos B\}\right]$

$=\frac{1}{k^{3}}\left[a^{2}\{b \cos B+c \cos C\}+2 b^{2}\{c \cos C+a \cos A\}+2 c^{2}\{a \cos A+a \cos B\}\right]$

$=\frac{1}{k^{3}}[a b(a \cos B+b \cos A)+b c(b \cos C+c \cos B)+a c(a \cos C+c \cos A)]$

$=\frac{1}{k^{3}}(a b c+b c a+a c b)$

$=3 a b c \times \frac{1}{k^{3}}$

$=3 \sin A \sin B \sin C \times \frac{1}{k^{3}} \times k^{3}$

$=3 \sin A \sin B \sin C$

= RHS

Hence proved.