The divergence of a vector field $\boldsymbol{u}=\pmatrix{u_1,u_2,u_3}$ is given by

$$\boldsymbol{\nabla\cdot u}=\frac{\partial u_1}{\partial x}+\frac{\partial u_2}{\partial y}+\frac{\partial u_3}{\partial z}.$$

a)

The variables $x$, $y$, and $z$ appear in a cyclical manner in each of the three components of $\boldsymbol{u}$.  Once you have calculated $\frac{\partial u_1}{\partial x}$, you can use cyclic permutations to determine the other two derivatives.  Hence

$$\begin{align}\frac{\partial u_1}{\partial x}&=\frac{\partial}{\partial x}\left(\simplify{({a1}*x+{b1}*y+{c1}*z)*({b1}*y-{c1}*z)}\right)\\&=\simplify{{a1}*({b1}*y-{c1}*z)},\end{align}$$

and so, cyclically permuting the variables,

$$\frac{\partial u_2}{\partial y}=\simplify{{a1}*({b1}*z-{c1}*x)}$$

and

$$\frac{\partial u_3}{\partial z}=\simplify{{a1}*({b1}*x-{c1}*y)}.$$

Finally, adding the components together gives the divergence

$$\boldsymbol{\nabla\cdot u}=\simplify[std]{{a1}*({b1}*y-{c1}*z)+{a1}*({b1}*z-{c1}*x)+{a1}*({b1}*x-{c1}*y)}=\simplify{{a1*b1-a1*c1}*x+{a1*b1-a1*c1}*y+{a1*b1-a1*c1}*z}.$$

b)

As in part a) the variables $x$, $y$, and $z$ appear cyclically in each component of $\boldsymbol{u}$, so we only need calculate one derivative explicitly, then use cyclic permutations to calculate the other two.  Hence

$$\begin{align}\frac{\partial u_1}{\partial x}&=\frac{\partial}{\partial x}\left\{\left(x^\var{p1}+y^\var{p1}\right)\left(\simplify{{a2}*y^{p1-1}-{a2}*z^{p1-1}}\right)\right\}\\&=\simplify{{p1}*x^{p1-1}}\left(\simplify{{a2}*y^{p1-1}-{a2}*z^{p1-1}}\right),\end{align}$$

and by symmetry

$$\frac{\partial u_2}{\partial y}=\simplify{{p1}*y^{p1-1}}\left(\simplify{{a2}*z^{p1-1}-{a2}*x^{p1-1}}\right),$$

and

$$\frac{\partial u_3}{\partial z}=\simplify{{p1}*z^{p1-1}}\left(\simplify{{a2}*x^{p1-1}-{a2}*y^{p1-1}}\right).$$

Finally, adding the derivatives together gives the divergence

$$\boldsymbol{\nabla\cdot u}=\simplify{{p1}*x^{p1-1}}\left(\simplify{{a2}*y^{p1-1}-{a2}*z^{p1-1}}\right)+\simplify{{p1}*y^{p1-1}}\left(\simplify{{a2}*z^{p1-1}-{a2}*x^{p1-1}}\right)+\simplify{{p1}*z^{p1-1}}\left(\simplify{{a2}*x^{p1-1}-{a2}*y^{p1-1}}\right)=0.$$

c)

First note that $f_1$ does not depend on $z$, $f_2$ does not depend on $y$, and $f_3$ does not depend on $z$.  This makes the differentiation very straight forward, and hence

$$\begin{align}\boldsymbol{\nabla\cdot u}&=\frac{\partial}{\partial x}\left(\simplify{{a3}*x}+f_1(y,z)\right)+\frac{\partial}{\partial y}\left(\simplify{{b3}*y}+f_1(x,z)\right)+\frac{\partial}{\partial z}\left(\simplify{{c3}*z}+f_1(x,y)\right)\\&=\simplify[all,!collectNumbers]{{a3}+{b3}+{c3}}\\&=\var{a3+b3+c3}.\end{align}$$