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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.

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    $\begingroup$ Try to group the terms and rewrite them going from lower order divided differences to higher order. E.g., to start, consider rewriting the term $f(x_0) - f(x_1)$ in terms of $f[x_0, x_1]$ and then continue the process $\endgroup$ – VorKir Mar 2 at 19:59
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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}

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