= Solution
A <principal body frame> is an orthonormal basis $\mathbf e_a(t)$ fixed in the rotating body and aligned with its principal axes. If $R(t)$ is the rotation matrix whose columns are these vectors, differentiating $R^TR=I$ shows that $\dot R R^T$ is <skew-symmetric>. Every skew-symmetric linear map in three dimensions is cross product with a unique vector $\boldsymbol\omega$, so
$$
\boxed{\dot{\mathbf e}_a=\boldsymbol\omega\mathbin\times\mathbf e_a.}
$$
For any vector $\mathbf A=A_a\mathbf e_a$,
$$
\left(\frac{d\mathbf A}{dt}\right)_{
m space}
=\dot A_a\mathbf e_a+\boldsymbol\omega\mathbin\times\mathbf A.
$$
Applying this to the conserved angular momentum
$$
\mathbf L=I_1\omega_1\mathbf e_1+I_2\omega_2\mathbf e_2+I_3\omega_3\mathbf e_3
$$
gives the <Euler equations for a torque-free rigid body>:
$$
\boxed{
I_1\dot\omega_1=(I_2-I_3)\omega_2\omega_3,
\quad
I_2\dot\omega_2=(I_3-I_1)\omega_3\omega_1,
\quad
I_3\dot\omega_3=(I_1-I_2)\omega_1\omega_2.}
$$
Multiplying these equations respectively by $\omega_i$ and summing proves conservation of <kinetic energy>; multiplying by $I_i\omega_i$ and summing proves conservation of squared <angular momentum>. Thus the <torque-free rigid-body invariants> are
$$
2E=I_1\omega_1^2+I_2\omega_2^2+I_3\omega_3^2,
\qquad
L^2=I_1^2\omega_1^2+I_2^2\omega_2^2+I_3^2\omega_3^2.
$$
Solving these two linear equations for the first two squared components gives
$$
\omega_1^2=
\frac{L^2-2EI_2+I_3(I_2-I_3)\omega_3^2}{I_1(I_1-I_2)},
$$
$$
\omega_2^2=
\frac{L^2-2EI_1+I_3(I_1-I_3)\omega_3^2}{I_2(I_2-I_1)}.
$$
Using $I_3\dot\omega_3=(I_1-I_2)\omega_1\omega_2$ yields the <quartic equation for one component of torque-free angular velocity>:
$$
\boxed{
\dot\omega_3^2=f(\omega_3)
=-\frac{
\left[L^2-2EI_2+I_3(I_2-I_3)\omega_3^2\right]
\left[L^2-2EI_1+I_3(I_1-I_3)\omega_3^2\right]
}{I_1I_2I_3^2}.}
$$
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