Help with tensor notation and curl

[stylophora]

Homework Statement

Show that $\nabla \times (a \cdot \nabla a) = a\cdot\nabla(\nabla\times a) + (\nabla \cdot a(\nabla \times a) - (\nabla \times a)\cdot \nabla a$

Homework Equations

$$\nabla \times (\nabla \phi) = 0$$
$$\nabla \cdot (\nabla \times a) = 0$$

The Attempt at a Solution

I started with breaking the LHS into two components. Observing that:
$$a\times(\nabla\times a) = 0.5 \nabla(a\cdot a) - a\cdot\nabla a$$

Taking the cross product of both sides:
$$\nabla\times a\times(\nabla\times a) = 0.5\nabla\times\nabla(a\cdot a) - \nabla\times(a\cdot \nabla a)$$

Recognizing that the term on the right corresponds to our initial equation.
$$\nabla \times (a \cdot \nabla a) = -\nabla\times a\times(\nabla\times a)$$

Unfortunately, I am sort of stuck here. One way that I have thought to go about it is by calling the LHS:
$$-\nabla\times a\times(\nabla\times a) = -\nabla\times c$$
where c = a\times(\nabla\times a)
I am confused on how to expand the above out using levi civita. I know that:
$$(\nabla \times c)_i = \epsilon_{ijk} \frac{dc_k}{dx_j}$$
But substituting in for c isn't making sense to me.

Sorry for the very rough attempt at a solution. I only started doing vector calc a week ago.

 

[mighty2000]

First, let's rewrite the LHS of the equation as:
\nabla \times (a \cdot \nabla a) = \nabla \times (\nabla \cdot (a a)) - \nabla \cdot (a \times (\nabla \times a))

Using the product rule for the first term on the RHS:
\nabla \times (\nabla \cdot (a a)) = (\nabla \cdot \nabla) (a a) + \nabla \cdot (\nabla a \times a)

Since the divergence of a gradient is always zero, the first term simplifies to zero. And using the vector identity \nabla \cdot (a \times b) = b \cdot (\nabla \times a) - a \cdot (\nabla \times b), we can rewrite the second term as:
\nabla \cdot (\nabla a \times a) = a \cdot (\nabla \times \nabla a) - \nabla a \cdot (\nabla \times a)

Substituting this back into our original equation:
\nabla \times (a \cdot \nabla a) = a \cdot (\nabla \times \nabla a) - \nabla a \cdot (\nabla \times a) - \nabla \cdot (a \times (\nabla \times a))

Using the vector identity \nabla \times \nabla a = 0, the first term on the RHS simplifies to zero. And using the fact that the divergence of a curl is always zero, the third term also simplifies to zero. This leaves us with:
\nabla \times (a \cdot \nabla a) = - \nabla a \cdot (\nabla \times a)

Now, let's focus on the RHS of the equation:
a \cdot \nabla (\nabla \times a) + (\nabla \cdot a) (\nabla \times a) - (\nabla \times a) \cdot \nabla a

Using the vector identity \nabla \times a = - (\nabla \times a), the third term on the RHS simplifies to zero. And using the fact