Please help solve this system of equations \left\{\begin{matrix}\frac{\sqrt[3]{2x+y}}{y}+\frac{\sqrt[3]{2x+y}}{2x}=(81)/(182) \\\frac{\sqrt[3]{2x-y}}{y}-\frac{\sqrt[3]{2x-y}}{2x…

Question

Please help solve this system of equations


`\left\{\begin{matrix}\frac{\sqrt[3]{2x+y}}{y}+\frac{\sqrt[3]{2x+y}}{2x}=(81)/(182) \\\frac{\sqrt[3]{2x-y}}{y}-\frac{\sqrt[3]{2x-y}}{2x}=(1)/(182) \end{matrix}\right.`

Answer 1

Make a substitution:

`\begin{cases}u=2x+y\\v=2x-y\end{cases}`

Then the system becomes

`\begin{cases}\frac{2\sqrt[3]{u}}{u-v}+\frac{2\sqrt[3]{u}}{u+v}=(81)/(182)\\\\\frac{2\sqrt[3]{v}}{u-v}-\frac{2\sqrt[3]{v}}{u+v}=\frac1{182}\end{cases}`

Simplifying the equations gives

`\begin{cases}\frac{4\sqrt[3]{u^4}}{u^2-v^2}=(81)/(182)\\\\\frac{4\sqrt[3]{v^4}}{u^2-v^2}=\frac1{182}\end{cases}`

which is to say,

`\frac{4\sqrt[3]{u^4}}{u^2-v^2}=\frac{81*4\sqrt[3]{v^4}}{u^2-v^2}`

`\implies\sqrt[3]{\left(\frac uv\right)^4}=81`

`\implies\frac uv=\pm27`

`\implies u=\pm27v`

Substituting this into the new system gives

`\frac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\frac1{182}\implies\frac1{v^2}=1\implies v=\pm1`

`\implies u=\pm27`

Then

`\begin{cases}x=\frac{u+v}4\\\\y=\frac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13`

(meaning two solutions are (7, 13) and (-7, -13))