[tex]\mathbf u+\mathbf v+\mathbf w=\mathbf 0[/tex]
Take the cross product of both sides with [tex]\mathbf u,\mathbf v,\mathbf w[/tex], respectively. By the distributive property of the cross product, you have the system
[tex]\begin{cases}\mathbf u\times\mathbf u+\mathbf u\times\mathbf v+\mathbf u\times\mathbf w=\mathbf 0\\\mathbf v\times\mathbf u+\mathbf v\times\mathbf v+\mathbf v\times\mathbf w=\mathbf 0\\\mathbf w\times\mathbf u+\mathbf w\times\mathbf v+\mathbf w\times\mathbf w=\mathbf 0\end{cases}\iff\begin{cases}\mathbf u\times\mathbf v+\mathbf u\times\mathbf w=\mathbf 0\\\mathbf v\times\mathbf u+\mathbf v\times\mathbf w=\mathbf0\\\mathbf w\times\mathbf u+\mathbf w\times\mathbf v=\mathbf 0\end{cases}[/tex]
since [tex]\mathbf x\times\mathbf x=\mathbf0[/tex] for any vector [tex]\mathbf x[/tex].
And because the cross product is anticommutative, you have
[tex]\begin{cases}\mathbf u\times\mathbf v-\mathbf w\times\mathbf u=\mathbf 0\\-\mathbf u\times\mathbf v+\mathbf v\times\mathbf w=\mathbf0\\\mathbf w\times\mathbf u-\mathbf v\times\mathbf w=\mathbf 0\end{cases}[/tex]
from which it follows immediately that
[tex]\mathbf u\time\mathbf v=\mathbf v\times\mathbf w=\mathbf w\times\mathbf u[/tex]
