I have the following derivation in my physics book I don't know how did they derive them
$\frac{d}{dt} \Sigma{_i}[(\vec{r}_{cm} + \vec{r_i})\times m_i(\vec{v}_{cm} + \vec{v_i})]$ = $\Sigma{_i}[\vec{v}_{cm} \times m_i \vec{v}_{cm} + \vec{r}_{cm} \times m_i \vec{a}_{cm} + \vec{v}_{cm}\times m_i\vec{v_i} + \vec{r}_{cm}\times m_i\vec{a_i} + \vec{v_i}\times m_i\vec{v}_{cm} + \vec{r_i}\times m_i\vec{a}_{cm} + \vec{v_i}\times m_i\vec{v_i} + \vec{r}\times m_i\vec{a_i}]$
I don't know how they expanded this cross product I understand how they derived the rest that is how they got $\tau= \vec{r}_{cm}\times \vec{a}_{cm} + \Sigma(\vec{r_i} \times F_i)$ which makes intuitive sense but I don't understand how they expanded that cross product in that particular manner if someone could shed some light on that it would be perfect.
Cross product is distributive over addition. $$(r_{cm} + r_i)\wedge (v_{cm} + v_{i}) = r_{cm}\wedge v_{cm} + r_{cm} \wedge v_{i} + r_{i}\wedge v_{cm} + r_i \wedge v_i $$ All these terms are time dependant so when you introduce the differential you must apply the product rule. So we get $$\dot{r}_{cm}\wedge v_{cm} +r_{cm}\wedge \dot{v}_{cm} + \dot{r}_{cm} \wedge v_{i} + r_{cm} \wedge \dot{v}_{i} + \dot{r}_{i}\wedge v_{cm} + r_{i}\wedge \dot{v}_{cm} + \dot{r}_i \wedge v_i +r_i \wedge \dot{v}_i $$ Mass is constant with time so you can just pull it out the derivate and cross product. The dots denote time derivatives so $\dot{r}_{cm} =v_{cm}$ etc.