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$xy^3z^3 = yx^3z^3 = zx^3y^3 \\ x^2+y^2+z^2$

I was wondering if the only solution to this was $z=y=x$

Solving an equation when there is a constraint is different right?

here there is not an infinite number of solution, but like only a dozen right?

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What's the constraint? Was there supposed to be a $\leq$ or $=$ on the second line? –  Karolis JuodelÄ— Nov 15 '12 at 5:53
    
The set of equations (without the constraint-but you don't indicate what $x^2+y^2+z^2 should be) were solved in this question where it was shown that another solution had one variable $0$. –  Ross Millikan Nov 15 '12 at 6:00
    
More than a dozen. With constraint say $x^2+y^2+z^2=1$, there are infinitely many with $z=0$. And another infinite bunch with $y=0$, and another with $x=0$, plus $8$ others. –  André Nicolas Nov 15 '12 at 6:24
    
yeah but i only need the extreme cases right? –  Arnold Nov 15 '12 at 16:38

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