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- Feb 14, 2012
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Find all the real solutions $(x, y, z)$ of the system:
$xyz=8$
$x^2y+y^2z+z^2x=73$
$x(y-z)^2+y(z-x)^2+z(x-y)^2=98$
$xyz=8$
$x^2y+y^2z+z^2x=73$
$x(y-z)^2+y(z-x)^2+z(x-y)^2=98$
Hi Jester, thanks for participating and you got it right!My solution
Expanding the third equation and using the first to eliminate $xyz$ and then using the second equation gives
$(x-y)(y-z)(z-x) = 0$
Clearly we cannot have $x = y= z$ as this violates the third equation. We haave three choices here. Let us pick $z = x$, the other follow
Thus
$x^2y = 8$ and $x^2y+xy^2+x^3=73$ or $xy^2 + x^3 = 65$ Now $y = \dfrac{8}{x^2}$ giving
$x^3 - 65 + \dfrac{64}{x^3}=0$ or $\dfrac{\left(x^3-1\right)\left(x^3-64\right)}{x^3} = 0$
Thus, $x = 1$ or $x = 4$ leading to $(1,8,1)$ and $(4,1/2,4)$.
The other solutions are obtained by interchanging these vlues.