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

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

Answer 1

Answer: (A)

$0$

As $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$
${\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C=3\sin A\sin B\sin C$
$\Rightarrow a^3+b^3+c^3=3abc$
Now $\Delta =\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=(a+b+c)\left|\begin{array}{ccc}1& b& c\\ 1& c& a\\ 1& a& b\end{array}\right|$
Applying $C_1\to C_1+C_2+C_3$
$\Delta =(a+b+c)\left|\begin{array}{ccc}1& b& c\\ 0& c-b& a-c\\ 0& a-b& b-c\end{array}\right|$
Applying $R_2\to R_2-R_1,R_3\to R_3-R_1$
$\Delta =(a+b+c)[(c-b)(b-c)-(a-b)(a-c)]=(a+b+c)[ab+bc+ca-a^2-b^2-c^2]=-(a^3+b^3+c^3)+3abc$
$\therefore \Delta =0$