Show that this vector identity is true using divergence theorem:

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Show that $$ \int_{V'}{\pmb{\nabla}' \times \frac{\pmb{F}(\pmb{r}')}{|\pmb{r}-\pmb{r}'|} }\,dV' \equiv \int_{S'}{\pmb{n}'\,\times\frac{\pmb{F}(\pmb{r}')}{|\pmb{r}-\pmb{r}'|}}\,dS'$$ where $\pmb{n}'$ is the outward pointing unit vector of the surface S' that bounds V', and $\pmb{F}(\pmb{r})$ is a 3-dimensional vector-valued function.

Hint: the integral conversion involves the Divergence Theorem (https://en.wikipedia.org/wiki/Divergence_theorem).

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Hint It seems to me that if u is a constant vector one has $$\nabla\cdot (G\times u) = u\cdot(\nabla\times G)$$ and $$n\cdot (G\times u) = u\cdot(n\times G)$$