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$$\sqrt[3]{x}=\sqrt[5]{32}$$
why can't i just cube both sides, therefore kick the x out of the cubic root, and then i would have
$$x = \sqrt{32} \\then \\x = \sqrt{16 * 2}\\x = 4\sqrt{2}$$

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$$x = 8$$

Step-by-step explanation:

$${x}^{ \frac{1}{3} } = {32}^{ \frac{1}{5} }$$

$${( {x}^{ \frac{1}{3} } )}^{3} = {( {32}^{ \frac{1}{5} } )}^{3}$$

$${x}^{1} = \sqrt[5]{ {32}^{3} }$$

$$x = \sqrt[5]{32768}$$

$$x = 8$$
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