<s>[INST] prove statement. cyclic sum complex numbers If $(a+b)^3 = (b+c)^3 = (c+a)³\,\,\,\,a,b,c\,\,\in\mathbb{C},\,\,a\neq b\neq c$, show that $a^3 = b^3 = c^3$.
Some hints would be great.
 [/INST] As $\displaystyle a\ne c, a+b\ne b+c\implies \frac{a+c}{b+c}\ne1$
So, from $\displaystyle(a+b)^3 = (b+c)^3 = (c+a)^3\implies a+b=\omega(b+c)=\omega^2(c+a)$ where $\omega$ is one of the two complex cube root of unity
From   $\displaystyle a+b=\omega(b+c),a+b(1-\omega)-c \omega=0\ \ \ \ (1)$
From  $\displaystyle\omega(b+c)=\omega^2(c+a),-a\omega^2+b\omega+c(\omega-\omega^2)=0\ \ \ \ (2)$
Solve for $a,b$ in terms of $c,$ then take the  cubes
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