 QUESTION

# In a $\Delta ABC$, a, b, c are the sides of the triangle opposite to the angles A,B,C, respectively.Then, the value of ${a^3}\sin \left( {B - C} \right) + {b^3}\sin \left( {C - A} \right) + {c^3}\sin \left( {A - B} \right)$ is equal to?A. 0B. 1C. 3D. 2

Hint: Since a, b, c are sides of a triangle, we will take the first term and find the summation of it, the other two terms will be similar to the first one.

$\sum {{a^3}\sin \left( {B - C} \right)}$
Using the formula$\left[ {\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = k} \right]$, we get,
$\Rightarrow \sum {{k^3}{{\sin }^3}A\sin \left( {B - C} \right)}$
$\Rightarrow \sum {{k^3}\left[ {{{\sin }^2}A\sin \left( {B + C} \right)\sin \left( {B - C} \right)} \right]}$
$\Rightarrow\sum {{k^3}\left[ {\left\{ {{{\sin }^2}A \times \dfrac{1}{2}\left( {\cos 2C - \cos 2B} \right)} \right\}} \right]}$
$\Rightarrow\sum {\dfrac{{{k^3}}}{2}\left[ {{{\sin }^2}A\left( {1 - 2{{\sin }^2}C - 1 + 2{{\sin }^2}B} \right)} \right]}$
$\Rightarrow \sum {\dfrac{{{k^3}}}{2}\left[ {{{\sin }^2}A - 2{{\sin }^2}A{{\sin }^2}C + 2{{\sin }^2}A{{\sin }^2}B - {{\sin }^2}A} \right]}$
$\Rightarrow \sum {\dfrac{{{k^3}}}{2}\left( 0 \right) = 0}$