= Solution
Expanding the symmetric spatial stencil by the <Taylor theorem> gives
$$
L_hu=\frac{\alpha+2\beta+2\gamma}{h^2}u+(\beta+4\gamma)u_{xx}+\frac{h^2}{12}(\beta+16\gamma)u_{xxxx}+\frac{h^4}{360}(\beta+64\gamma)u_{xxxxxx}+O(h^6).
$$
<Consistency of a numerical method> and cancellation of the $h^2$ term require $\alpha+2\beta+2\gamma=0$, $\beta+4\gamma=1$, and $\beta+16\gamma=0$. Solving this <linear system> gives
$$
\boxed{\alpha=-\frac52,\qquad\beta=\frac43,\qquad\gamma=-\frac1{12}}.
$$
The resulting <fourth-order centered second derivative> has
$$
L_hu=u_{xx}-\frac{h^4}{90}u_{xxxxxx}+O(h^6).
$$
The nonzero next coefficient proves that \b[the highest spatial <order of a numerical method> is four] for this three-parameter stencil.
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