First step is to add x to both sides.
Second step is cube both sides.
Solving this equation for x initially yields three possible solutions.
[tex]\begin{gathered} (4x)^{\frac{1}{3}}-x=0 \\ (4x)^{\frac{1}{3}}=x \\ 4x=x^3 \\ x^3-4x=0 \\ x(x^2-4)=0 \\ then_{} \\ x=0 \\ or \\ x^2-4=0 \\ x^2=4 \\ x=\pm2 \end{gathered}[/tex]Checking the solutions show that -2, 0 and 2 are valid solutions.