Simplify this expression with divided differences. 
The divided differences are defined as follows
  $$
f[x_i]
:= f(x_i), \quad
f[x_0, \ldots, x_n]
:= \frac{f[x_1, \ldots, x_n] - f[x_0, \ldots, x_{n - 1}]}{x_n - x_0} \quad \text{for } n \ge 2
$$
  For pairwise different $x_0,x_1,x_2$ simplify the expression
  $$
f(x_0) + f[x_0,x_1](x - x_0) + f[x_0,x_1,x_2](x - x_0)(x - x_2) - f(x_1) - f[x_1,x_2](x - x_1) - f[x_1,x_2,x_0](x - x_1)(x - x_2)
$$

I know that the $n$-th interpolation polynomial is given by
$$
 P_n(x)
 = f[x_0] + (x - x_0)f[x_0, x_1] + \ldots + \prod_{j = 0}^{n - 1} (x - x_j) f[x_0, \ldots, x_n].
$$
and hoped that the expression would maybe reduce to the second interpolation polynomial.
I am also aware of the mean value theorem for divided differences, but that didn't help either.
Then , I tried to utilise that the divided differences are invariant under permutations of the $x_i$ (namely $f[x_0,x_1,x_2] = f[x_1,x_2,x_0]$), so some simplification could be made, but I suspect that is not enough.
Lastly, I plain tried to write the divided differences in terms of $f(x_i)$ and $x_j$ but no cancelling was possible.
Note: This is one of 11 questions from a past exam for which one has 120 minutes and it only gives 2 from 40 possible points, and therefore there has to be a quick solution.

Taking @VorKir's hint:
We have
$$
f(x_0) - f(x_1)
= f[x_0,x_1](x_0 - x_1)
$$
and therefore
$$
f(x_0) + f[x_0,x_1](x - x_0) - f(x_1)
= f[x_0,x_1](x - x_1).
$$
Now,
\begin{align}
f[x_0,x_1](x - x_1) - f[x_1, x_2](x - x_1)
& = (x - x_1) \left( f[x_0,x_1] - f[x_1, x_2] \right) \\
& = (x - x_1)(x_0 - x_2) f[x_0,x_1,x_2],
\end{align}
since
$$
f[x_0,x_1,x_2] 
= \frac{f[x_1,x_2] - f[x_0,x_1]}{x_2 - x_0}.
$$
We have now reduced the expression down to
$$
(x - x_1)(x_0 - x_2) f[x_0,x_1,x_2] + f[x_0,x_1,x_2](x - x_0)(x - x_2) - f[x_1,x_2,x_0](x - x_1)(x - x_2).
$$
We now use that divided differences are invariant under permutations of the $x_i$ , so we have
$$
f[x_0,x_1,x_2]
= f[x_1,x_2,x_0]
$$
and therefore the expression is equal to
\begin{align}
& \ f[x_0,x_1,x_2]\left((x - x_1)(x_0 - x_2) + (x - x_0)(x - x_2) -(x - x_1)(x - x_2)\right) \\
= & \ f[x_0,x_1,x_2]\left( (x - x_1)(x_0 - x_2) + (x_1 - x_0)(x - x_2) \right) \\
= & \ f[x_0,x_1,x_2](x_0 - x) (x_2 - x_1).
\end{align}
