Tensor algebra, divergence of cross product

Hi there. I wanted to demonstrate this identity which I found in a book of continuum mechanics:



##curl \left ( \vec u \times \vec v \right )=div \left ( \vec u \otimes \vec v - \vec v \otimes \vec u \right ) ##



I've tried by writting both sides on components, but I don't get the same, I'm probably making some mistake.



##\displaystyle curl \left ( \vec u \times \vec v \right )= \epsilon_{ijk} \frac{\partial}{\partial x_j}(\epsilon_{kji}u_j v_i)= \epsilon_{ijk} \left [ \epsilon_{kji} \frac{\partial u_j}{\partial x_j}v_i +u_j \epsilon_{kji} \frac{\partial v_i}{\partial x_j} \right ] ##



I've used that ##(\vec u \times \vec v )_k=\hat e_k \cdot \epsilon_{ijk} u_j v_k \hat e_i=\delta_{ki} \epsilon_{ijk} u_j v_k=\epsilon_{kji}u_j v_i## I'm not sure if this is right.



Then, using the epsilon delta identity I've got ##rot \left ( \vec u \times \vec v \right )=6 \left [ \frac{\partial u_j}{\partial x_j}v_i +u_j \frac{\partial v_i}{\partial x_j} \right ]=6 \left [ (div \vec u) \vec v + (grad \vec v) \vec u \right ]##



Expanding the right hand side of the identity in components:



##div \left ( \vec u \otimes \vec v - \vec v \otimes \vec u \right ) = \frac{\partial}{\partial x_j}(u_i v_j -v_i u_j)=v_j \frac{\partial u_i}{\partial x_j}+ u_i \frac{\partial v_j}{\partial x_j} -u_j \frac{\partial v_i}{\partial x_j}- v_i \frac{\partial u_j}{\partial x_j}=(grad \vec u) \vec v+\vec u div \vec v - (grad \vec v) \vec u- \vec v div \vec u##



PD: The title should be curl of cross product instead of divergence. Sorry.