= Solution
For a vector $\mathbf A=\sum_i A_i\mathbf e_i$ in the <principal body frame>, the <derivative of a body-fixed basis vector> is $(d\mathbf e_i/dt)_{\widehat S}=\boldsymbol\omega\times\mathbf e_i$. Differentiating the expansion gives the <rotating-frame derivative formula>, also called the Coriolis formula:
$$
\left(\frac{d\mathbf A}{dt}\right)_{\widehat S}
=\left(\frac{d\mathbf A}{dt}\right)_S+\boldsymbol\omega\times\mathbf A.
$$
Apply this to <angular momentum> $\mathbf L=\sum_iI_i\omega_i\mathbf e_i$. The <principal moments of inertia> are constant in the <principal body frame>, so external <torque> balance becomes
$$
\boldsymbol\tau=\mathbf I\dot{\boldsymbol\omega}+\boldsymbol\omega\times(\mathbf I\boldsymbol\omega).
$$
Resolving the <cross product> gives <Euler's equations (rigid body dynamics)>:
$$
\boxed{\begin{aligned}
I_1\dot\omega_1+(I_3-I_2)\omega_2\omega_3&=\tau_1,\\
I_2\dot\omega_2+(I_1-I_3)\omega_3\omega_1&=\tau_2,\\
I_3\dot\omega_3+(I_2-I_1)\omega_1\omega_2&=\tau_3.
\end{aligned}}
$$
All components and dotted components in this display refer to the <principal body frame>.
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