= Solution
Direct multiplication gives
$$
AB=BA=\begin{pmatrix}0&10&-8\\-10&22&-10\\-8&10&0\end{pmatrix}.
$$
Take $D=B$ and $C=A$ in the preceding result: $B$ has the three distinct <eigenvalues> $2,4,6$ and the matrices commute. The common <eigenbasis> from part (i) gives
$$
\boxed{S=\begin{pmatrix}1&1&1\\1&0&2\\1&-1&1\end{pmatrix},\quad S^{-1}AS=\operatorname{diag}(1,2,2),\quad S^{-1}BS=\operatorname{diag}(2,4,6).}
$$
Indeed $AS=S\operatorname{diag}(1,2,2)$ and $BS=S\operatorname{diag}(2,4,6)$, and $\det S=2\ne0$. These equalities directly verify the <simultaneous diagonalization>.
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