Question

If for a triangle $ABC$, $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$, then ${\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C$ is equal to:

• A. $\sin A+\sin B+\sin C$
• B. $3\sin A\sin B\sin C$
• C. $\sin 3A+\sin 3B+\sin 3C$
• D. ${\sin }^{3}A{\sin }^{3}B{\sin }^{3}C$

Answer 1

The correct option is B $3\sin A\sin B\sin C$

$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0\Rightarrow (a+b+c)\left|\begin{array}{ccc}1& 1& 1\\ b& c& a\\ c& a& b\end{array}\right|=0(using{R}_1=R_1+R_2+R_3)$
$\Rightarrow (a+b+c)\left|\begin{array}{ccc}0& 0& 1\\ b-c& c-a& a\\ c-a& a-b& b\end{array}\right|=0$ (using $C_1=C_1-C_2,C_2=C_2-C_3$)
$\Rightarrow (a+b+c)[(b-c)(a-b)-{(c-a)}^2]=0$
$\Rightarrow (a+b+c)[ab-b^2-ac-+bc-c^2-a^2+2ac]=0$
$\Rightarrow -(a^3+b^3+c^3-3abc)=0$
$\Rightarrow (a^3+b^3+c^3-3abc)=0$
.
Now using the sine rule $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$, we get
$k^3({\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C-3\sin A\sin B\sin C)=0$
$\Rightarrow {\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C=3\sin A\sin B\sin C$