Question

If in a $△ABC,$
${\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C=3\sin A\sin B\sin C,$ then the value of the determinant $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$ is equal to

• A. $0$
• B. ${(a+b+c)}^3$
• C. $(a+b+c)(ab+bc+ca)$
• D. none of these

Answer 1

The correct option is A $0$

We have ${\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C=3\sin A\sin B\sin C$
Using sine rule in $\sin A=ak,\sin B=bk$ and $\sin C=ck$we get
$a^3{k}^3+b^3{k}^3+c^3{k}^3=3abc{k}^3$
$\therefore a^3+b^3+c^3=3abc$
$\Rightarrow a^3+b^3+c^3-3abc=0$
$\Rightarrow a+b+c=0$
$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$
$C\to C_1+C_2+C_3$
$=\left|\begin{array}{ccc}a+b+c& b+c+a& c+a+b\\ b& c& a\\ c& a& b\end{array}\right|$
$=(a+b+c)\left|\begin{array}{ccc}1& 1& 1\\ b& c& a\\ c& a& b\end{array}\right|$
Put $a+b+c=0$
$\therefore$ Value of determinant$=0\left|\begin{array}{ccc}1& 1& 1\\ b& c& a\\ c& a& b\end{array}\right|=0$