Question

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

• 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 $3\sin A\sin B\sin C$

$\Delta =\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=(a+b+c)\left|\begin{array}{ccc}1& b& c\\ 1& c& a\\ 1& a& b\end{array}\right|$
$[\text{applying}{C}_1\to C_1+C_2+C_3\text{and taking}(a+b+c\left)\text{common from}{C}_1\right]=(a+b+c)\left|\begin{array}{ccc}1& b& c\\ 0& c-b& a-c\\ 0& a-b& b-c\end{array}\right|[\text{applying}{R}_2\to R_2-R_1\text{and}{R}_3\to R_3-R_1]=(a+b+c)\left[\right(c-b\left)\right(b-c)-(a-b\left)\right(a-c\left)\right]=(a+b+c)[ab+bc+ca-a^2-b^2-c^2]=-(a^3+b^3+c^3)+3abc$
Now, $\Delta =0$ implies $a^3+b^3+c^3=3abc.$ Therefore, by the law of sines.
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$, we get
${\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C=3\sin A\sin B\sin C.$