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. 0
B. 1
C. 3
D. 2

Answer 1

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.

Complete step by step answer:
$\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} $
So, Option A is the correct answer.

Note: We started by finding the summation of the first term and finding the value of it, this value will be the same for the second and third term as well, therefore we can perform the summation once which saves our time.