<s>[INST] Find $\tan C$ in a triangle satisfying the constraint Given a triangle with angles $A,B, C$ and sides $a, b, c$ opposite to their respective angles, how can I find $\tan C$ such that
$$c^2={a^3+b^3+c^3\over a+b+c}$$
I used the law of Cosines on the LHS as well as on the cube terms but I don't know exactly I'm looking for in manipulating the expression. Should I try to get $\sin C$ and $\cos C$ separately, or is there a more elegant way of doing things? I believe there is since this is an old competition problem, I'm just not seeing it.
 [/INST] $$c^2={a^3+b^3+c^3\over a+b+c} \Rightarrow  c^2(a+b+c)=(a^3+b^3+c^3)$$
So $$c^2a+c^2b+c^3=a^3+b^3+c^3$$
$$\Rightarrow c^2a+c^2b=a^3+b^3=(a+b)(a^2-ab+b^2)$$
Since $a,b$ are non zero , $a+b \neq0$.
Thus $$c^2=(a^2-ab+b^2) \Rightarrow ab = a^2+b^2-c^2$$
$$\Rightarrow \frac{1}{2} = \frac{a^2+b^2-c^2}{2ab}$$
Thus $$\cos C = \frac{1}{2} $$
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