Solve:
$x^{2 / 3} + x^{1 / 3} - 2 = 0$

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Solution

We have,
$x^{2 / 3} + x^{1 / 3} - 2 = 0$
Let $x^{1 / 3} = y$
then,
$x^{2 / 3} + x^{1 / 3} - 2 = 0$
$y^{2} + y - 2 = 0$
$y^{2} + (2 - 1) y - 2 = 0$
$y^{2} + 2 y - 1 y - 2 = 0$
$y (y + 2) - 1 (y + 2) = 0$
$(y + 2) (y - 1) = 0$
If $y + 2 = 0$ , $y - 1 = 0$
$y = - 2$ , $y = 1$
$x^{1 / 3} = - 2$ , $x^{1 / 3} = 1$
$x = (- 2)^{3}$ , $x = 1^{3}$
$x = - 8$ , $x = 1$
Hence, this is the answer.

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