Given:
[tex](4x)^{\frac{1}{3}}-x=0[/tex]
To solve for x:
The first step in solving the equation is to add x on both sides, we get
[tex]\begin{gathered} (4x)^{\frac{1}{3}}-x+x=x \\ (4x)^{\frac{1}{3}}=x \end{gathered}[/tex]
The second step in solving the equation is to take the cube on both sides we get,
[tex]\begin{gathered} 4x=x^3 \\ x^3-4x=0 \end{gathered}[/tex]
Solving this equation for x initially yields three possible values.
[tex]\begin{gathered} x(x^2-4)=0\Rightarrow x=0^{} \\ Since,x^2=4 \\ \Rightarrow x=\pm2 \end{gathered}[/tex]
Hence, 0, -2, and 2 are the valid solutions.
