Say I have \begin{align*} p\equiv & p(\textbf{x},t),\text{ pressure}\\ \textbf{e}\equiv& \text{ any fixed vector}\\ \rho\equiv& \rho(\textbf{x},t),\text{ mass density}\\ \textbf{u}\equiv& \textbf{u}(\textbf{x},t),\text{ the velocity of the fluid particle} \end{align*}
I have to show that $$-\operatorname{div}(p\textbf{e}+\rho\textbf{u}(\textbf{u}\cdot\textbf{e}))=-\operatorname{div}(\rho\textbf{u})\textbf{u}\cdot\textbf{e}-\rho(\textbf{u}\cdot\nabla)\textbf{u}\cdot\textbf{e}-(\nabla p)\cdot\textbf{e}$$
Now what I have done so far is \begin{align*} &-\operatorname{div}(p\textbf{e}+\rho\textbf{u}(\textbf{u}\cdot\textbf{e}))\\ =& -\operatorname{div}(p\textbf{e})-\operatorname{div}(\rho\textbf{u}(\textbf{u}\cdot\textbf{e}))\\ =&-p\operatorname{div}(\textbf{e})-(\nabla p)\cdot\textbf{e}-\rho(\textbf{u}\cdot\textbf{e})\operatorname{div}(\textbf{u})-\nabla(\rho(\textbf{u}\cdot\textbf{e}))\cdot\textbf{u}\\ =&-p\operatorname{div}(\textbf{e})-(\nabla p)\cdot\textbf{e}-\rho(\textbf{u}\cdot\textbf{e})\operatorname{div}(\textbf{u})-(\textbf{u}\cdot\textbf{e})\textbf{u}\cdot\nabla\rho-\rho\textbf{u}\cdot\nabla(\textbf{u}\cdot\textbf{e})\\ =&-p\operatorname{div}(\textbf{e})-(\nabla p)\cdot\textbf{e}-(\textbf{u}\cdot\textbf{e})(\rho \operatorname{div}(\textbf{u})+\textbf{u}\cdot\nabla\rho)-\rho\textbf{u}\cdot\nabla(\textbf{u}\cdot\textbf{e})\\ =&-p\operatorname{div}(\textbf{e})-(\nabla p)\cdot\textbf{e}-(\textbf{u}\cdot\textbf{e})\operatorname{div}(\rho\textbf{u})-\rho\textbf{u}\cdot\nabla(\textbf{u}\cdot\textbf{e}) \end{align*}
If everything is done correctly thus far, then it seems it's almost there.
Is the first term above zero? Also is the last term above same as the second term of what I need to show?