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. 
 A: 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. 
A: $$a^3+b^3+c^3-3abc=(a+b)^3-3ab(a+b)+c^3-3abc=\{(a+b)+c\}\{(a+b)^2-(a+b)c+c^2\}-3ab\{(a+b)+c\}$$
$$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
But $2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2$
Now if $a^3+b^3+c^3-3abc=0$
either $a+b+c=0$
But with $A+B+C=m\pi,\tan A+\tan B+\tan C=\tan A\tan B\tan C$
As $0<A,B,C<\pi; \tan A\tan B\tan C\ne0$
So, we need $(\tan A-\tan B)^2+(\tan B-\tan C)^2+(\tan C-\tan A)^2=0\ \ \ \ (1)$
As $\tan A,\tan B,\tan C$ are real, each of $(\tan A-\tan B)^2,(\tan B-\tan C)^2,(\tan C-\tan A)^2$ must be $\ge0$
$(1)\implies$  $$(\tan A-\tan B)^2=(\tan B-\tan C)^2=(\tan C-\tan A)^2=0$$
Observation : The proposition will hold if in the condition, $\tan$ is replaced with $\sin,\cos$
Can you prove them?
