Solve the following equation.
$(x + y)^{2 / 3} + 2 (x - y)^{2 / 3} = 3 (x^{2} - y^{2})^{1 / 3}, 3 x - 2 y = 13$ .

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Solution

Dividing the first equation by $(x^{2} - y^{2})^{1 / 3}$ , we get
${(\frac{x + y}{x - y})}^{1 / 3} + 2 {(\frac{x - y}{x + y})}^{1 / 3} = 3$
Put ${(\frac{x + y}{x - y})}^{1 / 3} = u$ . Then $u + \frac{2}{u} = 3$ or $u^{2} - 3 u + 2 = 0$
or $(u - 1) (u - 2) = 0$ or $u = 1$ or $2$ .
Hence ${(\frac{x + y}{x - y})}^{1 / 3} = 1$ or $2$
First take $x + y = x - y$ , that is, $y = 0$
Then from the second equation
$3 x - 2 y = 13$
we get $x = 13 / 3$ . $(1)$
Again ${(\frac{x + y}{x - y})}^{1 / 3} = 2$ or $x + y = 8 x - 8 y$
or $7 x - 9 y = 0$ . $(2)$
Solving $(1)$ and $(2)$ , we get $x = 9, y = 7$
Hence the solutions are:
$x = 13 / 3 y = 0$ or $x = 9, y = 7$ .

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