Question

Let $A+B+C=\pi$ and $\alpha ={\sin }^3(B+C)\cdot \sin (2C+A),\beta ={\sin }^3(A+C)\cdot \sin (2A+B),\gamma ={\sin }^3(A+B)\cdot \sin (2B+C)$
are roots of the cubic equation $x^3+a{x}^2+bx+c=0$, then the value of $a$ is

• A. $0$
• B. $\sin A\sin B\sin C$
• C. $1$
• D. $\cos A\cos B\cos C$

Answer 1

The correct option is A $0$

$\alpha ={\sin }^3(B+C)\cdot \sin (2C+A)={\sin }^3(\pi -A)\cdot \sin (2C+\pi -B-C)={\sin }^{3}A\cdot \sin (B-C)={\sin }^{2}A\sin A\sin (B-C)={\sin }^{2}A\sin (B+C)\sin (B-C)={\sin }^{2}A({\sin }^{2}B-{\sin }^{2}C)={\sin }^{2}A({\sin }^{2}B-{\sin }^{2}C)$
Similarly,
$\beta ={\sin }^{2}B({\sin }^{2}C-{\sin }^{2}A)\gamma ={\sin }^{2}C({\sin }^{2}A-{\sin }^{2}B)$

Now,
$-a=\alpha +\beta +\gamma ={\sin }^{2}A({\sin }^{2}B-{\sin }^{2}C)+{\sin }^{2}B({\sin }^{2}C-{\sin }^{2}A)+{\sin }^{2}C({\sin }^{2}A-{\sin }^{2}B)\therefore -a=0\Rightarrow a=0$


Alternate solution:
If we assume $A=B=C={60}^{\circ }$, we get
$\alpha ={\sin }^3{120}^{\circ }\sin {180}^{\circ }=0$
Similarly,
$\beta =0,\gamma =0$
Sum of roots
$\alpha +\beta +\gamma =-a\Rightarrow 0=-a\therefore a=0$