= Solution
Use coordinates $\mathbf r$ measured from the <center of mass> $G$, so $\int\mathbf r\,dm=0$. The displacement from $P$ is $\mathbf r-\mathbf X$. Expand the defining <inertia tensor>:
$$
I_{ij}(P)=\int\left[|\mathbf r-\mathbf X|^2\delta_{ij}-(r_i-X_i)(r_j-X_j)\right]dm.
$$
All terms linear in $\mathbf r$ integrate to zero. The remaining constant terms give the <tensor parallel-axis theorem>,
$$
\boxed{I_{ij}(P)=I_{ij}(G)+M\left(|\mathbf X|^2\delta_{ij}-X_iX_j\right).}
$$
For the uniform cube, let $M=8\rho a^3$. Symmetry makes the off-diagonal <inertia tensor> entries zero. Direct integration gives $\int x_i^2\,dm=Ma^2/3$ for each coordinate, and therefore $I(G)=(2Ma^2/3)I_3$. With $\mathbf X=a(1,1,1)^T$, the <tensor parallel-axis theorem> gives
$$
\boxed{I(P)=Ma^2\begin{pmatrix}8/3&-1&-1\\-1&8/3&-1\\-1&-1&8/3\end{pmatrix}
=Ma^2\left(\frac{11}3I_3-\mathbf1\mathbf1^T\right).}
$$
The unit <eigenvector> $u=(1,1,1)^T/\sqrt3$ has <eigenvalue> $2Ma^2/3$. Every vector in the plane $x+y+z=0$ is annihilated by $\mathbf1\mathbf1^T$ and hence has <eigenvalue> $11Ma^2/3$. One orthonormal <basis> of that eigenspace is
$$
v=\frac{(1,-1,0)^T}{\sqrt2},\qquad w=\frac{(1,1,-2)^T}{\sqrt6}.
$$
\b[The principal moments are $2Ma^2/3$ along the body diagonal and $11Ma^2/3$ twice in its perpendicular plane.] Any orthonormal pair in that plane supplies the remaining principal axes.
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