In $△ A B C$ prove that, $a^{3} sin (B - C) + b^{3} sin (C - A) + c^{3} sin (A - B) = 0$ .

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Solution

L.H.S $= a^{3} sin (B - C) + b^{3} sin (C - A) + c^{3} sin (A - B)$

As we know that $\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c} = k$ by sine rule

$cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}, cos B = \frac{c^{2} + a^{2} - b^{2}}{2 a c}, cos C = \frac{b^{2} + a^{2} - c^{2}}{2 a b}$

So, $a^{3} sin (B - C) = a^{3} [sin B cos C - cos B sin C]$

$= a^{3} [k b \times \frac{b^{2} + a^{2} - c^{2}}{2 a b} - \frac{c^{2} + a^{2} - b^{2}}{2 a c} \times k c]$

$= k a^{3} \times \frac{b^{2} + a^{2} - c^{2} - c^{2} - a^{2} + b^{2}}{2 a}$

$= k a^{2} (b^{2} - c^{2})$

So, $b^{3} sin (C - A) = b^{3} [sin C cos A - cos C sin A]$

$= b^{3} [k c \times \frac{b^{2} + c^{2} - a^{2}}{2 b c} - \frac{a^{2} + b^{2} - c^{2}}{2 a b} \times k a]$

$= k b^{3} \times \frac{b^{2} + c^{2} - a^{2} - a^{2} - b^{2} + c^{2}}{2 b}$

$= k b^{2} (c^{2} - a^{2})$

So, $c^{3} sin (A - B) = c^{3} [sin A cos B - cos A sin B]$

$= c^{3} [k a \times \frac{a^{2} + c^{2} - b^{2}}{2 a c} - \frac{b^{2} + c^{2} - a^{2}}{2 b c} \times k b]$

$= k c^{3} \times \frac{a^{2} + c^{2} - b^{2} - b^{2} - c^{2} + a^{2}}{2 c}$

$= k c^{2} (a^{2} - b^{2})$

Now adding all these we get
$= k [a^{2} (b^{2} - c^{2}) + b^{2} (c^{2} - a^{2}) + c^{2} (a^{2} - b^{2})]$

$= k [a^{2} b^{2} - a^{2} c^{2} + b^{2} c^{2} - b^{2} a^{2} + c^{2} a^{2} - c^{2} b^{2}]$

$= k [0]$

$= 0$

$=$ R.H.S

Hence proved

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