Please help solve this system of equations[tex]\left\{\begin{matrix}\frac{\sqrt[3]{2x+y}}{y}+\frac{\sqrt[3]{2x+y}}{2x}=\frac{81}{182} \\\frac{\sqrt[3]{2x-y}}{y}-\frac{\sqrt[3]{2x-y}}{2x}=\frac{1}{182} \end{matrix}\right.[/tex]

Make a substitution:[tex]\begin{cases}u=2x+y\\v=2x-y\end{cases}[/tex]Then the system becomes[tex]\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}[/tex]Simplifying the equations gives[tex]\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}[/tex]which is to say,[tex]\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}[/tex][tex]\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81[/tex][tex]\implies\dfrac uv=\pm27[/tex][tex]\implies u=\pm27v[/tex]Substituting this into the new system gives[tex]\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1[/tex][tex]\implies u=\pm27[/tex]Then[tex]\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13[/tex](meaning two solutions are (7, 13) and (-7, -13))