= Solution
Choose, for example,
$$
\boxed{
\mathbf B_0=(0,0,B_0),
\qquad
\mathbf E_0=(cB_0,0,0)},
\qquad B_0\ne0.
$$
The fields are perpendicular and satisfy $E_0^2=c^2B_0^2$, so both invariants vanish. Every nonzero constant background with vanishing invariants can be brought to this form by a spatial rotation, up to signs.
In this frame, and with a convention consistent with the <relativistic Lorentz force>, the mixed tensor is
$$
F^\mu{}_\nu
=B_0
\begin{pmatrix}
0&1&0&0\\
1&0&1&0\\
0&-1&0&0\\
0&0&0&0
\end{pmatrix}.
$$
Calling the displayed dimensionless matrix $N$, direct multiplication gives
$$
N^2=
\begin{pmatrix}
1&0&1&0\\
0&0&0&0\\
-1&0&-1&0\\
0&0&0&0
\end{pmatrix},
\qquad
N^3=0.
$$
Consequently
$$
\boxed{
F^\mu{}_\rho F^\rho{}_\sigma F^\sigma{}_\nu=0}.
$$
This is a tensor equation. Since it holds in the canonical frame, it holds in every Lorentz frame for any <null electromagnetic field>.
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