How to solve a difficult system Problem: solve in $\mathbb R^{3}$ this system
$$
\begin{cases}y(1+x+x^2+x^3)=\dfrac{z}{16}\\ y^2x(1+x+2x^2+x^3+x^4)=\dfrac{2z+17}{16}\\ 
y^3x^3(1+x+x^2+x^3)=\dfrac{z}{16}\\ 
y^4x^6=1
\end{cases}
$$
Wolfram alpha gives $(x,y,z)=\left(4,\dfrac{1}{8},170\right)$.
I don't know how to solve it, but my attempt is as follows: 
First equation: 
$1+x+x^2+x^3=\dfrac{z}{16y}$ 
So by equation $2$ we find:   
$y^{2}x\left(\dfrac{z}{16y}+x^2+x^4\right)=\dfrac{2z+17}{16}$
I don't know how to complete my work. 
 A: The fourth equation is redundant.  From the first and the third equation (by equating them and noting that $1+x+x^2+x^3=0$ iff $x=-1$, but $x=-1$ doesn't yield a real solution), you can already deduce that $y^2x^3=1$.  If you set $x=t^2$ and $y=1/{t^3}$, then the first and the second equations are now $$\frac{z}{16}=t^3+\frac{1}{t^3}+t+\frac{1}{t}=s(s^2-2)$$ and $$\frac{2z+17}{16}=t^4+\frac{1}{t^4}+t^2+\frac{1}{t^2}+2=(s^2-2)(s^2-1),$$
where $s=t+\frac{1}{t}$.  That is
$$(s^2-2)(s^2-1)-2s(s^2-2)=\left(\frac{2z+17}{16}\right)-2\left(\frac{z}{16}\right)=\frac{17}{16}.$$
Therefore
$$s^4-2s^3-3s^2+4s+\frac{15}{16}=0.$$
If $u=2s$, then
$$u^4-4u^3-12u^2+32u+15=0$$
or 
$$(u-5)(u+3)(u^2-2u-1)=0.$$
That is,
$$u=5,-3,1\pm\sqrt{2}.$$
By AM-GM, $u=2s=2\left(t+\frac1t\right)$ satisfies $|u|\geq 4$.  Thus, $u=5$ is the only solution.  Therefore, $s=5/2$, so that $t=2$ or $t=1/2$.
In any case, this means
$$z=16s(s^2-2)=16\left(\frac52\right)\left(\frac{25}{4}-2\right)=170.$$
If $t=1/2$, then
$$(x,y,z)=(1/4,8,170).$$
If $t=2$, then
$$(x,y,z)=(4,1/8,170).$$
So wolframalpha missed a solution.
In fact, all solutions (real or complex) apart from the two solutions above are
$$(x,y,z)=\left(\left(\frac{-3\pm i\sqrt7}{4}\right)^2,\left(\frac{-3\mp i\sqrt7}{4}\right)^3,-6\right)$$
$$(x,y,z)=\left(\left(\frac{1+\sqrt{2}\pm i \sqrt{13-2\sqrt{2}}}{4}\right)^2,\left(\frac{1+\sqrt{2}\mp i \sqrt{13-2\sqrt{2}}}{4}\right)^3,-7-4\sqrt2\right)$$
$$(x,y,z)=\left(\left(\frac{1-\sqrt{2}\pm i \sqrt{13+2\sqrt{2}}}{4}\right)^2,\left(\frac{1-\sqrt{2}\mp i \sqrt{13+2\sqrt{2}}}{4}\right)^3,-7+4\sqrt2\right).$$
