<s>[INST] If $\tan^3A+\tan^3B+\tan^3C=3\tan(A)\tan(B)\tan(C)$, prove triangle ABC is equilateral triangle If $\tan^3A+\tan^3B+\tan^3C=3\tan(A)\tan(B)\tan(C)$, prove triangle ABC is equilateral triangle
Now i remember a identity which was like if $a+b+c=0$,then $a^3+b^3+c^3=3abc$. So i have $\sum_{}^{} \tan(A)=0$. How do i proceed?
Thanks. 
 [/INST] Note that if $A,B,C$ are angles of a triangle, we have that $$\tan A+\tan B +\tan C=\tan A \tan B \tan C$$ As seen here. If $$0=\tan A+\tan B +\tan C=\tan A \tan B \tan C$$
Then we have that for at least one angle among $A,B,C$ is $0$ or $\pi$. Thus, it is a contradiction. So $$\tan A+\tan B +\tan C \neq 0$$
Now exploit the fact $$a^3+b^3+c^3-3abc=0$$ is true only if $a=b=c$ given that $a+b+c \neq 0$. Thus $\tan A=\tan B= \tan C$. Thus $\triangle {ABC}$ is equilateral. 
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