Question

In a $△ABC$, a,b and c are the side of the triangle opposite to the angles $A,B,C$ respectively. Then the value of $a^3\sin (B-C)+b^3\sin (C-A)+c^3\sin (A-B)$ is equal to:

• A. $0$
• B. $1$
• C. $3$
• D. $2$

Answer 1

The correct option is A $0$

We have: In $\Delta ABC$.

SIne law: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

$a^3\sin (B-C)+b^3\sin (C-A)+c^3\left(\sin (A-B)\right)$

$=a^3(\sin B.\cos C-\cos B.\sin C)+b^3(\sin C.\cos A-\cos C.\sin A)+c^3(\sin A.\cos B-\cos A.\sin B)$

$=a^3(\frac{b}{2R}.\cos C-\cos B\frac{c}{2R})+b^3(\frac{c}{2R}\cos A-\frac{a}{2R}\cos C)+c^3(\frac{a}{2R}\cos B-\frac{b}{2R}\cos A)$

$=\frac{\cos A}{2R}({cb}^3-c^{3}b)+\frac{\cos B}{2R}(c^{3}a-a^{3}c)+\frac{\cos C}{2R}(a^{3}b-b^{3}a)$

$=bc.\frac{\cos A}{2R}(b^2-c^3)+ac.\frac{\cos B}{2R}(c^2-a^2)+\frac{\cos C}{2R}.ab(a^2-b^2)$

$=\frac{1}{4R}[(b^2+c^2-a^2)(b^2-c^2)+(a^2+c^2-b^2)(c^2-a^2)+(a^2+b^2-c^2)(a^2-b^2)]=0$

Hence;

$a^3\sin (B-C)+b^3\sin (C-A)+c^3\sin (A-B)=0$