$x^2 y = 1$, $y^2 z = 128$, $z^2 x = 32$ I tried to do this simultaneous equation in my additional maths class at school, and not even the teacher could do it using elimination method (how we were asked to do it). 
Through trial and error, we got the answer $x = 1/2$, $y = 4$, $z = 8$. However, we couldn’t find a way to solve it algebraically.
Is this possible to do by elimination, or just by substitution? Could the responde include both methods if so?
Thanks in advance
 A: HINT
\begin{align*}
(x^{2}y)(y^{2}z)(z^{2}x) = (xyz)^{3} = 2^{12} \Longleftrightarrow xyz = 16
\end{align*}
Therefore the system of equations proposed is equivalent to
\begin{cases}
z = 16x\\
y = 8x\\
z = 2y\\
xyz = 16
\end{cases}
A: Since any solution has to be given by positive numbers we are free to assume $x=2^a, y=2^b, z=2^c$ and solve 
$$ \left\{\begin{array}{rcl}2a+b&=&0\\ 2b+c&=&7 \\ 2c+a&=& 5\end{array}\right. $$
leading to $(a,b,c)=(-1,2,3)$ and $(x,y,z)=\left(\frac{1}{2},4,8\right)$.
A: The other hints are fine, but there are inevitably alternative approaches. None of $x, y, z$ can be zero, so we are fine multiplying and dividing.
Take $y^2z=128$ and multiply by $x^4$ so that $x^4y^2z=128x^4$. But $x^4y^2=(x^2y)^2=1$ so $$z=128x^4$$
Then substitute in the final equation to obtain $$(128x^4)^2x=32$$which you can solve for $x$ and hence find $y$ and $z$ too.
A: Hint:
$$x^2y=1\Longrightarrow y=\dfrac{1}{x^2}$$
$$y^2z=128\Longrightarrow \left( \dfrac{1}{x^2}\right)^2 z=128\Longrightarrow z=128x^4$$
$$z^2x=32 \Longrightarrow (128x^4)^2x=32 \Longrightarrow x^9=\dfrac{32}{128^2}=\dfrac{1}{2^9}$$
This gives $x=\dfrac{1}{2}$.
A: Hint: 
$$x^3y^3z^3 = 2^7\cdot 2^5= 2^{12}\implies xyz=  2^4 = 16$$
A: By elimination:
$(1)$ gives $x^2 = \frac{1}{y}$
Put this into $(2)$ and you get:
$\frac{z}{x^4} = 128$ or $z = 128x^4.$
Putting this into $(3)$ and you arrive at $x^9128^2 = 32 \implies x^9 = \frac{1}{512}.$
Notice that $2^9 = 512$, so $x = \frac{1}{2}$ which you can then sub in to find the other values.
A: Hint: Use logarithms to turn the system into a linear system, which is probably more familiar.
Let $a=\log_2 x$, $b=\log_2 y$, $c=\log_2 z$. Then
$$
2a+b=0,
\quad
2b+c=7,
\quad
2c+a=5
$$
whose solution is
$$
a = -1, \quad
b = 2, \quad
c = 3
$$
This gives
$$
x = 2^{-1}, \quad
y = 2^2, \quad
z = 2^3
$$
