Show that if at least one of the four angles A ± B ± C is a multiple of π, then $$\sin^4A + \sin^4 B + \sin^4 C − 2 \sin^2 B \sin^2 C − 2 \sin^2 C \sin^2 A − 2 \sin^2 A \sin^2 B + 4 \sin^2 A \sin^2 B \sin^2 C = 0$$
I want to start with proving $\sin(A+B+C)$ or $(\sin(A)+\sin(B)+\sin(C))^2$, however, I failed in both cases.
On
Hint:
First of all writing $\sin A=a$ etc.,
$$a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=(a^2+b^2-c^2)^2-(2ab)^2$$
$$=(a+b+c)(a+b-c)(a-b+c)(a-b-c)$$
Now if $A+B+C=\pi$
by this $\sin A+\sin B+\sin C=4\cos\dfrac A2\cos\dfrac B2\cos\dfrac C2$
and by this $\sin A+\sin B-\sin C=4\sin\dfrac A2\sin\dfrac B2\cos\dfrac C2$
Use $\sin2x=2\sin x\cos x$
We shall same expressions in some order if $A\pm B\pm C=\pi$
the following information may help you towards a solution.
since all the sine terms are squared we may as well assume that A,B and C are the angles of a triangle. let the sides be $a,b,c$ in the usual configuration. then if $\Delta$ represents the area of the triangle we have the two relations: $$ \Delta = \sqrt{s(s-a)(s-b)(s-c)} $$ and $$ R = \frac{abc}{4\Delta} $$ where $s = \frac{a+b+c}2$ is the semi-perimeter and $R$ is the circumradius.
if you eliminate $\Delta$ and substitute for $s$ you have a polynomial relation in $a,b,c$ which will give you the required result after applying the sine rule: $$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R $$