If a body contains no planes of symmetry, the principal moments of inertia can be determined mathematically. To show how this is done, consider the rigid body which is spinning with an angular velocity $\omega$, directed along one of its principal axes of inertia. If the principal moment of inertia about this axis is $I$, the angular momentum can be expressed as $\mathbf{H}=l \omega=I \omega_{x} \mathbf{i}+I \omega_{y} \mathbf{j}+I \omega_{z} \mathbf{k}$. The components of $\mathbf{H}$ may also be expressed by Eqs. $21-10$, where the inertia tensor is assumed to be known. Equate the $\mathbf{i}, \mathbf{j},$ and $\mathbf{k}$ components of both expressions for $\mathbf{H}$ and consider $\omega_{x}, \omega_{y},$ and $\omega_{z}$ to be unknown. The solution of these three equations is obtained provided the determinant of the cocfficicnts is zero. Show that this determinant, when expanded, yields the cubic equation
$$\begin{array}{l}
t^{3}-\left(I_{x x}+I_{y y}+I_{z z}\right) I^{2} \\
\quad+\left(I_{x x} I_{y y}+I_{y y} I_{x z}+I_{x z} I_{x x}-I_{x y}^{2}-I_{y z}^{2}-I_{z x}^{2}\right) I \\
\quad-\left(I_{x x} I_{y y} I_{z z}-2 I_{x y} I_{y z} I_{z x}-I_{x x} I_{y z}^{2}-I_{y y} I_{z x}^{2}-I_{z z} I_{x y}^{2}\right)=0
\end{array}$$