Prove or disprove the implication: Prove or disprove the implication:
$a^2\cdot \tan(B-C)+ b^2\cdot \tan(C-A)+ c^2\cdot \tan(A-B)=0 \implies$
$ ABC$ is an isosceles triangle.
I tried to break down the left hand side in factors, but all efforts were in vain.
Does anyone have a suggestion? 
Thank you very much!

The claim holds if we replace the function $\tan(B-C)$ (and cyclic permutations) with $\tan^*(B-C)=\frac{\sin(B-C)}{\cos(B+C)}$:
Since:
$$\tan^*(B-C)=\frac{\sin B \cos C - \sin C \cos B}{\cos B \cos C - \sin B \sin C}$$
and $2ab\cos(C) = a^2+b^2-c^2, 2R\sin A=a, 2ab\sin C=4\Delta$, we have:
$$2R\tan^*(B-C)= \frac{2abc(a^2+b^2-c^2-a^2-c^2+b^2)}{(a^2+c^2-b^2)(a^2+b^2-c^2)-16\Delta^2},$$
but Heron's formula gives $16\Delta^2 = 4b^2c^2-(b^2+c^2-a^2)^2$, from which
$$2R\tan^*(B-C) = \frac{4abc(b^2-c^2)}{a^4-(b^2-c^2)^2-4b^2c^2+(b^2+c^2-a^2)^2}$$ follows. This gives:
$$2R\tan^*(B-C) = \frac{2abc(b^2-c^2)}{a^2(a^2-b^2-c^2)}=\frac{c^2-b^2}{a \cos A}$$ and:
$$2Ra^2\tan^*(B-C) = \frac{a(c^2-b^2)}{\cos A} = 2R(c^2-b^2)\tan(A),$$
$$ a^2 \tan^*(B-C) = (c^2-b^2)\tan A = (c^2-b^2)\frac{4\Delta}{b^2+c^2-a^2}.\tag{1}$$
So we have, in terms of side lengths,
$$ \sum_{cyc}(c^2-b^2)(a^2+c^2-b^2)(a^2+b^2-c^2) = 0, \tag{2} $$
or:
$$ \sum_{cyc}(c^2-b^2)(a^4-b^4-c^4+2b^2c^2) = 0,$$
$$ \sum_{cyc}\left(b^6-c^6-a^4b^2+a^2b^4+3b^2c^4-3b^4c^2\right)=0, $$
$$ \sum_{cyc}\left(-a^4b^2+a^2b^4+3b^2c^4-3b^4c^2\right)=0, $$
$$ \sum_{cyc}\left(4b^2c^4-4b^4c^2\right)=0,\tag{3} $$
Now it is easy to notice that $a=\pm b,a=\pm c$ or $b=\pm c$ imply that the LHS is zero, so $(a^2-b^2)(b^2-c^2)(c^2-a^2)$ divides the LHS in $(3)$. 
$$ \sum_{cyc}\left(4b^2c^4-4b^4c^2\right) = 4(a^2-b^2)(b^2-c^2)(c^2-a^2)\tag{4} $$
proves the (modified) claim.
