= Solution
Use the standard ordered <basis> $(\mathbf e_1,\mathbf e_2,\mathbf e_3)$ for both maps. The <reflection in a hyperplane> interchanges the second and third coordinates, so its <matrix> is
$$
\boxed{A=\begin{pmatrix}1&0&0\\0&0&1\\0&1&0\end{pmatrix}}.
$$
Orient the rotation axis along $\mathbf n=(1,1,1)/\sqrt3$ and use the right-hand convention. The <rotation matrix> for angle $2\pi/3$ sends $\mathbf e_1$ to $\mathbf e_2$, $\mathbf e_2$ to $\mathbf e_3$, and $\mathbf e_3$ to $\mathbf e_1$. For example, the <Rodrigues rotation formula> gives
$$
R\mathbf e_1=\cos\frac{2\pi}{3}\,\mathbf e_1+\sin\frac{2\pi}{3}(\mathbf n\times\mathbf e_1)+\left(1-\cos\frac{2\pi}{3}\right)(\mathbf n\cdot\mathbf e_1)\mathbf n=\mathbf e_2.
$$
Multiplying the <rotation matrix> by the dilation factor gives
$$
\boxed{B=2R=\begin{pmatrix}0&0&2\\2&0&0\\0&2&0\end{pmatrix}}.
$$
Direct multiplication gives $A^2=I$, $B^2=4\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}$ and $B^3=8I$. Hence \b[$B^3=8A^2$]. If the opposite axis orientation is chosen, $B=2R^T$ also has cube $8I$. Independently of coordinates, two applications of the <reflection> give the identity, while three applications of the rotation-dilation give a full turn and dilation by $2^3$. Every <basis> represents these resulting maps by $I$ and $8I$. Thus the equality even holds when the two <matrices> are expressed in different <bases>.
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