Question

In a triangle $ABC$,if $si{n}^{3}A+si{n}^{3}B+si{n}^{3}C=3sinAsinBsinC$

then $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=?$

• A. $0$
• B. $(a+b+c{)}^3$
• C. $(a+b+c)(ab+bc+ca)$
• D. $ab+bc+ca$

Answer 1

The correct option is A $0$

According to sine Rule,

$a=2RsinAb=2RsinBc=2RsinC$

Now,$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=abc-a^3-b^3+abc+abc-c^3\Rightarrow 3abc-(a^3+b^3+c^3)=8R^3[3sinAsinBsinC-(si{n}^{3}A+si{n}^{3}B+si{n}^{3}C\left)\right]$

and it is given that $3sinAsinBsinC=si{n}^{3}A+si{n}^{3}B+si{n}^{3}C$

$\Rightarrow 8R^3[3sinAsinBsinC-(si{n}^{3}A+si{n}^{3}B+si{n}^{3}C\left)\right]=8R^3\times 0=0$

Therefore, Answer is $A$