ep pls help

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Solve for $x$: $4x^{1/3}-2 \cdot \frac{x}{x^{2/3}}=7+\sqrt[3]{x}$.

Guest Jun 24, 2021

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Best Answer

let y = $\sqrt[3]{x}$ :

$4y-\frac{2y^3}{y^2}=7+y\\4y-2y=7+y\\y=7$

$7=\sqrt[3]{x}\\x=343$

textot Jun 24, 2021

textot · Answer 1 · 2021-06-24T04:05:12

let y = $\sqrt[3]{x}$ :

$4y-\frac{2y^3}{y^2}=7+y\\4y-2y=7+y\\y=7$

$7=\sqrt[3]{x}\\x=343$