Vector calculus problem I have to solve this:
$$[(\nabla \times \nabla)\cdot \nabla](x^2 + y^2 + z^2)$$
But I am really drowning in the sand..
Can anybody help me please?
 A: HINT:
We have a third ordered derivative acting on a quadratic term.
SPOILER ALERT:  SCROLL OVER THE SHADED AREA TO REVEAL SOLUTION

 Using a form of Einstein's tensor notation where summation is implied when indices are repeated, we have 
 \begin{align}(\hat x_i\times \hat x_j)\cdot \hat x_k(\partial_i\partial_j\partial_k)(x_{\ell}^2)&=(\hat x_i\times \hat x_j)\cdot \hat x_k(\partial_i\partial_j)2x_{\ell}\delta_{k\ell}\\\\&=(\hat x_i\times \hat x_j)\cdot \hat x_k(\partial_i)2\delta_{j\ell}\delta_{k\ell}\\\\&=0\end{align}

A: Nice question:
First of all we apply the gradient:
$$\nabla (x^2 + y^2 + z^2) = (2x,2y,2z).$$
Then lets evaluate the vector $\nabla \times \nabla$. Using the cross product we get:
$\nabla \times \nabla = (\hat{i}\partial_x + \hat{j}\partial_y + \hat{k}\partial_z)\times (\hat{i}\partial_x + \hat{j}\partial_y + \hat{k}\partial_z) = \hat{k}\partial_{xy} - \hat{j}\partial_{xz} - \hat{k}\partial_{yx} + \hat{i}\partial_{yz} - \hat{j}\partial_{zx} -\hat{i}\partial_{zy} = \hat{i}(\partial_{yz} - \partial_{zy}) + \hat{j}(\partial_{zx} - \partial_{xz}) + \hat{k}(\partial_{xy} - \partial_{yx}).$
Then, making the inner product:
$$\hat{i}(\partial_{yz} - \partial_{zy}) + \hat{j}(\partial_{zx} - \partial_{xz}) + \hat{k}(\partial_{xy} - \partial_{yx}) \cdot (2x,2y,2z) = 0. $$
