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$$ xy^3z^3 = yx^3z^3 = zx^3y^3 $$ Is there a way to solve this system?

I think the answer is 1, but i can't verify my intuition.

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up vote 1 down vote accepted

Clearly if one of the variables is $0$ then all three equations are satisfied. Suppose then that's not the case. We rewrite as $$y^2 = x^2,\ \ x^2 = z^2,\ \ y^2 = z^2$$ Letting $x$ be a free-variable, we have $y = \pm x$ and $z=\pm x$.

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So $\left(x,y,z\right) = x\left(1,\pm 1, \pm 1\right)$. – Eric Angle Nov 15 '12 at 5:08
what is a free variable? isn't the answer 1? – Arnold Nov 15 '12 at 5:09
Free variable means you are free to choose the value as you please. $1$ is not the only possible solution. – EuYu Nov 15 '12 at 5:10
and how do you get to y^2 = x^2 x^2 = z^2 and y^2 = z^2? – Arnold Nov 15 '12 at 5:11
Group the equations in pairs and divide out common variables. – EuYu Nov 15 '12 at 5:11

Another way of looking at it, clearly $x=0$, or $y=0$, or $z=0$ is a solution. So, starting with \begin{alignat*}{3} xy^3z^3 &= yx^3z^3 &&= zx^3y^3 \end{alignat*} And considering the case where $x \ne 0$, $y \ne 0$, $z \ne 0$ \begin{alignat*}{3} y^2z^2 &= x^2z^2 &&= x^2y^2 \end{alignat*} Considering each pair of equalities, we get: \begin{align*} y^2z^2 &= x^2z^2 & x^2z^2 &= x^2y^2 & y^2z^2 &= x^2y^2 \\ \implies y &= \pm x & z &= \pm x & z &= \pm x \\ \implies y &= \pm x & z &= \pm x \end{align*} Hence the solutions are points of the form $(0,a,b)$, $(a,0,b)$, $(a,b,0)$, or $(a, \pm a, \pm a)$ for $a,b \in \mathbb{R}$.

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