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 $a^3\sin (B-C)+b^3\sin (C-A)+c^3\sin (A-B)$ is equqal to

• A. 0
• B. 1
• C. 3
• D. 2

Answer 1

Answer: (A)

$\sum a^3\sin (B-C)$
$=\sum k^3{\sin }^{3}A\sin (B-C)$
$\left[\because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\right]$
$=\sum k^3\left[{\sin }^{2}A\sin (B+C)\sin (B-C)\right]$
$=k^3[\left\{{\sin }^{2}A\times \frac{1}{2}(\cos 2C-\cos 2B)\right\}+{\sin }^{2}B\times \frac{1}{2}(\cos 2A-\cos 2C)+{\sin }^{2}C\times \frac{1}{2}(\cos 2B-\cos 2A)]$
$=\frac{k^3}{2}[{\sin }^{2}A\left(1-2{\sin }^{2}C-1+2{\sin }^{2}B\right)+{\sin }^{2}B\left(1-2{\sin }^{2}A-1+2{\sin }^{2}C\right)+{\sin }^{2}C\left(1-2{\sin }^{2}B-1+2{\sin }^{2}A\right)]$
$=\frac{k^3}{2}[-2{\sin }^{2}A{\sin }^{2}C+2{\sin }^{2}A{\sin }^{2}B-2{\sin }^{2}B{\sin }^{2}A+2{\sin }^{2}B{\sin }^{2}C-2{\sin }^{2}C{\sin }^{2}B+2{\sin }^{2}C{\sin }^{2}A]$
$=\frac{k^3}{2}[0]=0$