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 will follow the method suggested by Blue in his comment to show that the implication does not hold.
The sides $a$, $b$, $c$ are proportional to $\sin A$, $\sin B$ and $\sin C$, respectively, by the sine law, so the equation is equivalent to $S = 0$, where
$$ S = \sin^2 A \tan(B-C) + \sin^2 B \tan(C-A) + \sin^2 C \tan(A - B).$$
Letting $x = \tan A$, $y = \tan B$, and $z = \tan C$, we can rewrite $S$ in terms of $x$, $y$, $z$; for example, we have $\sin^2 A = x^2/(1 + x^2)$ and $\tan(B - C) = (y-z)/(1 + yz)$. (We will ignore for now the special case where $ABC$ is a right triangle, and one of $x$, $y$, $z$, is correspondingly infinite.) Thus
$$ S = \frac{x^2}{1 + x^2} \frac{y-z}{1 + yz} + \frac{y^2}{1 + y^2} \frac{z-x}{1 + zx} + \frac{z^2}{1 + z^2} \frac{x-y}{1 + xy.}$$ 
Now we note that $S = 0$ when any two of $x, y, z$ are equal, so we expect the numerator of $S$ to be divisible by $(y-z)(z-x)(x-y)$. This is indeed the case (I used a computer for this part, although it is possible to do long division first by $y-z$, then by $z-x$ and finally by $x-y$):
$$S = \frac{(y-z)(z-x)(x-y)(2x^2 y^2 z^2 + x^2 y^2 + y^2 z^2 + z^2 x^2 - 1)}{(1 + yz)(1 + zx)(1 + xy)(1 + x^2)(1 + y^2)(1 + z^2)}.$$
Now that we have recovered the factors $y-z$, $z-x$ and $x - y$, we can split off factors equal to the tangents that originally appeared in $S$, so that 
$$S = \frac{y-z}{1 + yz} \frac{z-x}{1 + zx} \frac{x-y}{1 + xy} T = \tan(B-C)\tan(C-A)\tan(A-B) T,$$
where 
$$ T = \frac{2x^2 y^2 z^2 + x^2 y^2 + y^2 z^2 + z^2 x^2 - 1}{(1 + x^2)(1 + y^2)(1 + z^2)}.$$
To simplify $T$, we write $X = 1 + x^2$, $Y = 1 + y^2$, $Z = 1 + z^2$, so that $X = 1 + \tan^2 A = \sec^2 A$, and similarly, $Y = \sec^2 B$, $Z = \sec^2 C$. Then by replacing $x^2$, $y^2$ and $z^2$ with $X-1$, $Y-1$, $Z-1$, respectively, where they appear in $T$, we obtain
$$ 
\begin{align}
T &= \frac{2(X-1)(Y-1)(Z-1) + (X-1)(Y-1) + (Y-1)(Z-1) + (Z-1)(X-1) - 1}{XYZ} \\
  &= \frac{2XYZ - XY - YZ - ZX}{XYZ} \\
  &= 2 - \frac{1}{X} - \frac{1}{Y} - \frac{1}{Z} \\
  &= 2 - \cos^2 A - \cos^2 B - \cos^2 C.
\end{align}
$$
We therefore have the following identity:
$$S =  \tan(B-C)\tan(C-A)\tan(A-B) (2 - \cos^2 A - \cos^2 B - \cos^2 C),$$
which corresponds more or less to Blue's factorization. Interestingly, we never used the fact that $A$, $B$ and $C$ were the angles of a triangle. The identity extends to the case where one of the angles is a right angle by continuity.
Now the only question becomes whether it is possible to find a scalene triangle for which
$$\cos^2 A + \cos^2 B + \cos^2 C = 2.$$ 
If we assume that $A < B < C$, then this amounts to showing that the function 
$$f(A,B) = \cos^2 A + \cos^2 B + \cos^2 (A + B)$$
takes the value $2$ somewhere in the interior of the triangle $\Delta$ bounded by the lines $A = 0$, $B = A$ and $A + 2B = \pi$. At the vertex $(0,0)$ of $\Delta$, we find $f(0,0) = 3$, so by continuity, at points in the interior of $\Delta$ near $(0,0)$, we must have $f(A,B) \approx 3$. Likewise, at the vertex corresponding to an equilateral triangle we have $f(\pi/3, \pi/3) = 3/4$, so at some points in the interior of $\Delta$, we must have $f(A, B) \approx 3/4$. Since the interior of $\Delta$ is a connected set, by the intermediate value theorem, we must have $f(A, B) = 2$ for some choice of $(A, B)$ in the interior of $\Delta$, which therefore corresponds to a scalene triangle.
Question: Was it predictable somehow, before the factoring step, that the numerator of $T$ would be expressible in terms of $x^2$, $y^2$ and $z^2$? That is what allowed the method to work as smoothly as it did.
