Solve the system $x^2(y+z)=1$ ,$y^2(z+x)=8$ and $z^2(x+y)=13$ Solve the system of equations in real numbers 
\begin{cases} x^2(y+z)=1 \\  y^2(z+x)=8 \\z^2(x+y)=13 \end{cases}
My try:
Equations can be written as:
\begin{cases}\frac{1}{x}=xyz\left(\frac{1}{y}+\frac{1}{z}\right)\\
\frac{8}{y}=xyz\left(\frac{1}{x}+\frac{1}{z}\right)\\
\frac{13}{z}=xyz\left(\frac{1}{y}+\frac{1}{x}\right)\end{cases}
Let $p=xyz.$ Then we have:
\begin{cases}\frac{1}{x}-\frac{p}{y}-\frac{p}{z}=0\\
\frac{p}{x}-\frac{8}{y}+\frac{p}{z}=0\\
\frac{p}{x}+\frac{p}{y}-\frac{13}{z}=0\end{cases}
Then we get
$$\frac{p+1}{x}=\frac{p+8}{y}=\frac{p+13}{z}$$
Any clue here?

Let $a=1/x, b=1/y, c=1/z$.
$$a-pb-pc=0$$
$$pa-8b+pc=0$$
$$pa+pb-13c=0$$
From here, you can get $$(p+1)a=(13+p)c\\\text{and}$$
$$(p+1)a=(p+8)b$$
Substitute those values for $b$ and $c$ in last equation to get
$$p^3+11p^2-52=0$$
It'll give you p, and then you can find the corresponding values of $x,y$ and $z$.
