Answer
\begin{array}{l}
x^{4}-x^{2} y^{2}+x+y\\
x y(y z+1)
\end{array}
Work Step by Step
Given:
\[
x y z+x=f(x, y, z)
\]
(a) From $13.1 .1,$ we know that we have to calculate $c$, substituting $x, y$ and $z$ in the function
\[
\begin{aligned}
\because x y z+x=f(x, y, z) & \\
(x+y)(x-y) x^{2}+x+y =f\left(x+y, x-y, x^{2}\right) &\\
x^{2}\left(x^{2}-y^{2}\right)+x+y =f\left(x+y, x-y, x^{2}\right) &\\
x^{4}-x^{2} y^{2}+x+y=f\left(x+y, x-y, x^{2}\right) &
\end{aligned}
\]
(b) Similarly substituting,
\[
\begin{aligned}
x y * \frac{y}{x} * x z+x y=f\left(x y, \frac{y}{x}, x z\right) & \\
x y(y z+1)=f(-3,2,1) &
\end{aligned}
\]
