<s>[INST] If $(\sin^{-1}x)^3 + (\sin^{-1}y)^3 + 3\sin^{-1}x\sin^{-1}y = 1$ which of the following would be true If $(\sin^{-1}x)^3 + (\sin^{-1}y)^3 + 3\sin^{-1}x\sin^{-1}y = 1$  which of the following is true:
A) $\frac{\sin ^{-1} x}{1+\cos ^{-1} y}=\frac{\pi}{2}$
B) $\frac{x+y}{\sin 1}=-2$
C) $\sin ^{-1} x+\sin ^{-1} y=1$
D) $\frac{x+y}{\cos 1}=1$
My Approach: Well I couldn't do really much. I tried to consider approaches related to factorising the given expression but that doesn't work in this case. How should I try to approach this problem? Any hints/solutions are appreciated. Thanks!
 [/INST] Consider the Gauss identity
\begin{align}
a^3+b^3+c^3-3abc &= (a+b+c)(a^2+b^2+c^2-ab-ac-bc) \\
& = (a+b+c) \frac{1}{2} ((a-b)^2+(a-c)^2+(b-c)^2). 
\end{align}
If $a^3+b^3+c^3-3abc=0$, plugin $a=\sin^{-1}(x)$, $b=\sin^{-1}, c=-1$ we have the hypothesis and consequently two cases.
Case 1 $a+b+c=0$ then
$\sin^{-1}(x)+ \sin^{-1}(y)-1=0$ which is C).
Case 2 $(a-b)^2+(a-c)^2+(b-c)^2= 0$ which implies $a=b=c=0$ then $\sin^{-1}(x)=\sin^{-1}(y)=-1$ but this is not possible for all $x,y \in \mathbb{R}$ so the only posibility is C).
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