= Solution
Place the origin at the centre and the axes along the edges. Symmetry makes the off-diagonal entries of the <inertia tensor> zero. Uniform density gives $\langle x^2\rangle=a^2/12$, $\langle y^2\rangle=b^2/12$, $\langle z^2\rangle=c^2/12$. Consequently the <principal moments of inertia> are
$$
\boxed{I_1=\frac M{12}(b^2+c^2),\quad I_2=\frac M{12}(c^2+a^2),\quad I_3=\frac M{12}(a^2+b^2).}
$$
For example $I_1=\rho\int(y^2+z^2)\,dx\,dy\,dz$ gives the first expression directly.
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