Solve the system $x^2(y+z)=1$ ,$y^2(z+x)=8$ and $z^2(x+y)=13$

Question

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?

Answer 1

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 $z$.

Answer 2

I would substitute $$y+z=a,z+x=b,x+y=c$$ then $$z=\frac{a+b-c}{2}$$ and so on. Eliminating the variables $$y,z$$ we get for $x$: $$25 x^9-3828 x^6+109 x^3=-42$$

Answer 3

Alternatively, denote $y=ax, z=abx$. Then: $$\begin{cases} x^2(y+z)=1 \\ y^2(z+x)=8 \\z^2(x+y)=13 \end{cases} \Rightarrow \begin{cases} ax^3(1+b)=1 \\ a^2x^3(1+ab)=8 \\a^2b^2x^3(1+a)=13 \end{cases}.$$ Divide $(2)$ by $(1)$ and $(3)$ by $(2)$: $$\begin{cases} a(ab+1)=8(1+b) \Rightarrow b=\frac{8-a}{a^2-8} \\ 8b^2(1+a)=13(1+ab) \Rightarrow 8\cdot \frac{(8-a)^2}{(a^2-8)^2}\cdot (1+a)=13(1+a\cdot \frac{8-a}{a^2-8}) \Rightarrow \end{cases}\\ 8(8-a)^2(1+a)=13\cdot 8\cdot (a-1)(a^2-8) \Rightarrow \\ (3a-10)(2a^2+7a-2)=0.$$ You can find $a$, then $b$, then $x$, $y$ and $z$.