Show that $a^{3} c o s (B - C) + b^{3} (C - A) + c^{3} c o s (A - B) = 3 a b c .$

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Solution

$a^{3} c o s (B - C) + b^{3} c o s (C - A) + c^{3} c o s (A - B)$

= $k^{3} s i n^{3} A c o s (B - C) + k^{3} s i n^{3} B c o s (C - A) + k^{3} (A - B)$

= $k^{3} [s i n^{2} A s i n (B + C) c o s (B - C) + s i n^{2} C s i n (A + B) c o s (A - B)]$

= $\frac{1}{2} k^{3} [s i n^{2} A (s i n 2 B + s i n 2 C) + s i n^{2} B (s i n 2 C + s i n 2 A) + s i n^{2} C (s i n 2 A + s i n 2 B)]$

= $k^{3} [s i n^{A} s i n B c o s B + s i n^{2} A s i n C c o s C + s i n^{2} B s i n C c o s C + s i n^{2} B s i n A c o s A + s i n^{2} C s i n A c o s A + s i n A c o s B]$

= $k^{3} [s i n A s i n B (s i n A c o s B + c o s A s i n B) + s i n B s i n C (s i n B c o s C + c o s B s i n C) + s i n C s i n A (s i n C s i n A (s i n C c o s A + c o s C s i n A)]$

= $k^{3} [s i n A s i n B s i n (A + B) + s i n B s i n C s i n (B + C) + s i n C s i n A s i n (C + A)$

= $k^{3} [s i n A s i n B s i n C + s i n B s i n C s i n A + s i n C s i n A s i n B]$

$= 3 k^{3} s i n A . k s i n B . k s i n C = 3 a b c .$
LHS=RHS
Hence proof

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