Vector calculus identity for $\nabla\times(\vec{b}\cdot\nabla)\vec{b}$

I'm going through a paper on turbulence and in it the author uses the following $$\nabla\times(\vec{b}\cdot\nabla)\vec{b}=(\vec{b}\cdot\nabla)(\nabla\times\vec{b})-\left((\nabla\times\vec{b})\cdot\nabla\right)\vec{b}$$ however I have tried to verify this with both vector analysis identities and using suffix notation and I can't seem to do so. I wondered if anyone could either show why it holds or correct it. Thanks in advance.

• Do you know how to compute the curl of a vector field and the divergence of a vector field? I'm assuming $\vec{b}$ is your given vector field. – MathNewbie May 22 '15 at 11:09
• Sorry I meant I used both methods, as in manipulating standard vector identities and using suffix notation. The specific identity I tried to apply was $$\nabla\times(\psi\vec{A})=\psi\nabla\times\vec{A}+\nabla\psi\times\vec{A}$$ since $\vec{b}\cdot\nabla$ is a scalar this seemed the most appropriate. However this didn't yield the correct result for me. – Nick Bell May 22 '15 at 11:12
• Yes I do know how to compute the curl and the divergence and yes b is my vector. But it is an algebraic quantity and so I do not have numeric values for the vector so can't do a direct calculation. – Nick Bell May 22 '15 at 11:14
• Well $\vec{b}$ should be a vector field. Not sure what you mean by an algebraic quantity, but write out $\vec{b}$ in terms of its three components. – MathNewbie May 22 '15 at 11:17
• I realise that you could do that to verify the equivalence. But I was more hoping for an explanation of how you get from the left hand side to the right hand side using standard vector manipulation. eg identities, product rules etc. – Nick Bell May 22 '15 at 11:27

EDIT: It turns out the identity is incorrect. Consider the vector field $\vec{b}=\begin{pmatrix}xy\\0\\xy\end{pmatrix}$ (it doesn't work)