<s>[INST] If $x\sin A+y\sin B+z\sin C=x^2\sin2A+y^2\sin2B+z^2\sin 2C=0$, show $x^3\sin3A+y^3\sin3B+z^3\sin 3C=0$ I'm having trouble with a question that came in one of my exam and is about complex numbers and trigonometry:

If $$x \sin A +y \sin B+z\sin C=0$$ and $$x^2 \sin 2A + y^2 \sin 2B + z^2 \sin 2C=0$$ then prove that $$x^3 \sin 3A+y^3 \sin 3B +z^3 \sin 3C=0$$ where $x$, $y$, $z$ belong to $\mathbb{R}$ and $A+B+C=\pi$.

What I tried: I tried maybe doing it as imaginary part of $xe^{iA} + ye^{iB} +ze^{iC} =0$ and similarly, but I'm reaching a dead end in that approach
 [/INST] If $\sin{C}=0$, so $$x\sin{A}+y\sin(180^{\circ}k-A)=0,$$ where $k$ is an integer number.
Thus, easy to show (just consider two cases: $k$ is odd and $k$ is even) that $$x^3\sin3A+y^3\sin(540^{\circ}k-3A)=0.$$
Now, let $\prod\limits_{cyc}\sin{A}\neq0.$
Thus, the first condition gives $$z=-\frac{x\sin{A}+y\sin{B}}{\sin(A+B)}$$ and with the second we obtain:
$$(x^2\sin2A+y^2\sin2B)\sin(A+B)=2(x\sin{A}+y\sin{B})^2\cos(A+B)$$ or
$$(\sin{A}\cos{A}\sin(A+B)-\sin^2A\cos(A+B))x^2-2\sin{A}\sin{B}\cos(A+B)xy+$$
$$+(\sin{B}\cos{B}\sin(A+B)-\sin^2B\cos(A+B))=0$$ or
$$\sin{A}\sin{B}(x^2-2xy\cos(A+B)+y^2)=0$$ or
$$x^2-2xy\cos(A+B)+y^2=0,$$ which since $|\cos(A+B)|\neq1,$ gives $$x=y=0.$$
Can you end it now?
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