Question

Solve the following system of equations.
$\frac{x}{y^2}+\frac{y}{x^2}=\frac{9}{8},{\log }_{2}x+{\log }_{\sqrt{2}}\sqrt{y}=3$

Answer 1

$\frac{x}{y^2}+\frac{y}{x^2}=\frac{9}{8}$

$x^3+y^3=\frac{9}{8}{\left(xy\right)}^2\to 1$

${\log }_{2}x+{\log }_{\sqrt{2}}\sqrt{y}=3$

${\log }_{2}x+{\log }_{2}y=3$

${\log }_{2}x=3$

$\left(xy\right)=2^3=8$

$\left(xy\right)=8\to 2$

From $1$ and $2$

$x^3+{\left(\frac{8}{x}\right)}^3=\frac{9}{8}{\left(8\right)}^2$

$x^6+{\left(8\right)}^3=\left(72\right)x^3$

$x^6-72x^3+{\left(8\right)}^3=0$

$x^3=\frac{72\pm \sqrt{{\left(72\right)}^2-4{\left(8\right)}^3}}{2}=\frac{72\pm \sqrt{{\left(56\right)}^2}}{2}$$x^3=\frac{72\pm 56}{2}=64,8$

$x=2,4$

From $2$: $xy=8$

$y=\frac{8}{x}y=\frac{8}{2}=4,y=\frac{8}{4}=2$

$(x,y):(2,4),(4,2)$