Question

Solve $(x+y{)}^{\frac{2}{3}}+2(x-y{)}^{\frac{2}{3}}=3(x^2-y^2{)}^{\frac{1}{3}}....(1)$.
$3x-2y=13.....\left(2\right)$.

Answer 1

${(x+y)}^{\frac{2}{3}}+2{(x-y)}^{\frac{2}{3}}=3{(x^2-y^2)}^{\frac{1}{3}}.........\left(1\right)3x-2y=\mathrm{13..........}\left(2\right)$

From $\left(1\right)$,

${(x+y)}^{\frac{2}{3}}+2{(x-y)}^{\frac{2}{3}}=3{(x^2-y^2)}^{\frac{1}{3}}\Rightarrow {(x+y)}^{\frac{2}{3}}+2{(x-y)}^{\frac{2}{3}}=3{\left(\right(x+y\left)\right(x-y\left)\right)}^{\frac{1}{3}}\Rightarrow \frac{{(x+y)}^{\frac{2}{3}}}{{\left(\right(x+y\left)\right(x-y\left)\right)}^{\frac{1}{3}}}+2\frac{{(x-y)}^{\frac{2}{3}}}{5{\left(\right(x+y\left)\right(x-y\left)\right)}^{\frac{1}{3}}}=3$

$\Rightarrow {\left(\frac{x+y}{x-y}\right)}^{\frac{1}{3}}+2{\left(\frac{x-y}{x+y}\right)}^{\frac{1}{3}}=3$

Let ${\left(\frac{x+y}{x-y}\right)}^{\frac{1}{3}}=t$ then ${\left(\frac{x-y}{x+y}\right)}^{\frac{1}{3}}=\frac{1}{t}$

$\Rightarrow t+\frac{2}{t}=3t^2-3t+2=0\Rightarrow t^2-2t-t+2=0\Rightarrow t(t-2)-1(t-2)=0\Rightarrow (t-1)(t-2)=0t=1,2$

But ${\left(\frac{x+y}{x-y}\right)}^{\frac{1}{3}}=t$

$\Rightarrow \frac{x+y}{x-y}=t^3$

For $t=1$

$\frac{x+y}{x-y}=1x+y=x-y\Rightarrow y=\mathrm{0.......}\left(3\right)$

For $t=2$

$\frac{x+y}{x-y}=8x+y=8x-8y\Rightarrow 7x=9y.........\left(4\right)$

Now substituting $\left(3\right)$ in $\left(2\right)$
we get $x=\frac{13}{3}$

$\Rightarrow (\frac{13}{3},0)$ is a solution

Solving $\left(4\right)$ and $\left(2\right)$

we get $x=9$ and $y=7$

$\Rightarrow (9,7)$ is the other solution