jmc

To start, note that $\frac{x}{x^{2/3}}=x^{1-\frac{2}{3}}=x^{1/3}$. Also note that we can rewrite the cube root with a fractional exponent, so $\sqrt[3]{x}=x^{1/3}$. Using these pieces of information, rewrite the given equation as: $$4x^{1/3}-2x^{1/3}=7+x^{1/3}$$ Move all $x^{1/3}$ terms to one side and simplify: $$\begin{array}{rl}2x^{1/3}-x^{1/3}& =7\\ \Rightarrow x^{1/3}& =7\\ \Rightarrow (x^{1/3}{)}^3& =7^3\\ \Rightarrow x& =\boxed{343}\end{array}$$