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- Feb 14, 2012

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$xyz=8$

$x^2y+y^2z+z^2x=73$

$x(y-z)^2+y(z-x)^2+z(x-y)^2=98$

- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

- 3,812

$xyz=8$

$x^2y+y^2z+z^2x=73$

$x(y-z)^2+y(z-x)^2+z(x-y)^2=98$

- Jan 26, 2012

- 183

$(x-y)(y-z)(z-x) = 0$

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.

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- Feb 14, 2012

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Hi Jester, thanks for participating and you got it right!

$(x-y)(y-z)(z-x) = 0$

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.

To be completely honest with you, my solution is tedious compared to yours and hence I don't think it's worth mentioning here...