= Solution
Write $\boldsymbol r_i=\boldsymbol X+\boldsymbol r_i'$ and $\boldsymbol v_i=\boldsymbol V+\boldsymbol v_i'$, where $\boldsymbol V=\dot{\boldsymbol X}$. The definition of the <centre of mass> and its <derivative> give
$$
\sum_i m_i\boldsymbol r_i'=0,\qquad
\sum_i m_i\boldsymbol v_i'=0.
$$
Expand the total <angular momentum>:
$$
\begin{aligned}
\boldsymbol L
&=\sum_i m_i(\boldsymbol X+\boldsymbol r_i')\times(\boldsymbol V+\boldsymbol v_i')\\
&=M\boldsymbol X\times\boldsymbol V
+\boldsymbol X\times\sum_i m_i\boldsymbol v_i'
+\left(\sum_i m_i\boldsymbol r_i'\right)\times\boldsymbol V
+\sum_i m_i\boldsymbol r_i'\times\boldsymbol v_i'.
\end{aligned}
$$
The two middle terms vanish. Thus the <angular momentum decomposition about the centre of mass> is
$$
\boxed{\boldsymbol L=M\boldsymbol X\times\boldsymbol V
+\sum_i m_i\boldsymbol r_i'\times\boldsymbol v_i'.}
$$
The first term is orbital <angular momentum> of the centre; the second is <angular momentum about the centre of mass>.
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