If $x^{2 / 3} + x^{1 / 3} - 2 = 0$ then $x =$

- 2, 1

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- 8, - 2

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- 8, 1

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none of these

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Solution

The correct option is D $- 8, 1$
$(\sqrt[3]{x})^{2} + \sqrt[3]{x} - 2 = 0$
$\Rightarrow$ $(\sqrt[3]{x})^{2} + 2 \sqrt[3]{x} - \sqrt[3]{x} - 2 = 0$
$\Rightarrow$ $\sqrt[3]{x} (\sqrt[3]{x} + 2) - 1 (\sqrt[3]{x} + 2) = 0$
$\Rightarrow$ $(\sqrt[3]{x} + 2) (\sqrt[3]{x} - 1) = 0$
$\Rightarrow$ $\sqrt[3]{x} + 2 = 0$ and $\sqrt[3]{x} - 1 = 0$
$\Rightarrow$ $\sqrt[3]{x} = - 2$ and $\sqrt[3]{x} = 1$
Cubing both sides,
$\Rightarrow$ We get, $x = - 8$ and $x = 1$

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