MHB Solve x,y,z: Real Solutions $(x,y,z)$

[anemone]

Find all the real solutions $(x, y, z)$ of the system:

$xyz=8$

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

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

 

[Mathwebster]

My solution

Spoiler

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.

 

[anemone]

My solution

Spoiler

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.

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

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