 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:All terms linear in $\mathbf{r}$ integrate to zero. The remaining constant terms give the tensor parallel-axis theorem,For the uniform cube, let $M=8\rho a^3$. Symmetry makes the off-diagonal inertia tensor entries zero. Direct integration gives $\int x_i^{2}dm=M{a}^2/3$ for each coordinate, and therefore $I(G)=(2M{a}^2/3)I_3$. With $\mathbf{X}=a(1,1,1{)}^T$, the tensor parallel-axis theorem givesThe unit eigenvector $u=(1,1,1{)}^T/\sqrt{3}$ has eigenvalue $2M{a}^2/3$. Every vector in the plane $x+y+z=0$ is annihilated by $11^T$ and hence has eigenvalue $11M{a}^2/3$. One orthonormal basis of that eigenspace isThe principal moments are $2M{a}^2/3$ along the body diagonal and $11M{a}^2/3$ twice in its perpendicular plane. Any orthonormal pair in that plane supplies the remaining principal axes.