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!

I believe I have an example in which the identity is true but the triangle is not isosceles.  All the algebra in the following was done by Maple.
Let $a=1$ and $b=2$, and take $c^2$ to be a root of the cubic $x^3-5x^2-25x+45$.  The cubic has two roots between $1$ and $9$, either of which will give a valid set of sides $\{a,b,c\}$ for a triangle.
Now $c^2\ne4$ since $4$ is not a root of the cubic; so $c\ne2$; likewise $c\ne1$; so the triangle is not isosceles.  Future forbidden values of $c$ are checked in the same way and will be given without comment.
Now use the cosine rule to find $\cos A$, $\cos B$ and $\cos C$ in terms of $a,b,c$.  Then use the sine rule to write $\sin A$ and $\sin B$ as multiples of $\sin C$.  Check that $\cos C\ne\pm1$ and so $\sin C\ne0$.  Check that $\cos(B-C)$ and so on are not zero, then calculate $\tan(B-C)$ and so on, replacing $\sin^2C$ by $1-\cos^2C$; they all have a factor of $\sin C$.
Form the given expression and cancel $\sin C$; after simplification, Maple gives the numerator of the result as a constant times
$$(c-1)(c-2)(c+1)(c+2)(c^6-5c^4-25c^2+45)\ ,$$
which is zero.
This also fits in with the answer given by @Blue: with the values of $a,b,c$ I have specified, the first factor in his/her product is zero and not the others.
