Solve for $x$: $4x^{1/3}-2 \cdot \frac{x}{x^{2/3}}=7+\sqrt[3]{x}$.
Note that \(x^{1/3}=\frac{x}{x^{2/3}}=\sqrt[3]x\) so let \(u=x^{1/3}\) and we have \(4u-2u=7+u\)
Simplifying, we get \(u=7\) so \(x=u^3=7^3=343\)
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