Answer :
The given equation is
[tex](4x)^{\frac{1}{3}}-x=0[/tex]The first step, add x to both sides
[tex]\begin{gathered} (4x)^{\frac{1}{3}}-x+x=0+x \\ (4x)^{\frac{1}{3}}=x \end{gathered}[/tex]Cube each side ------ 2nd answer
[tex]\begin{gathered} (4x)^{\frac{1}{3}\times3}=x^{1\times3} \\ 4x=x^3 \end{gathered}[/tex]Now, Subtract 4x from both sides
[tex]\begin{gathered} 4x-4x=x^3-4x \\ 0=x^3-4x \\ x^3-4x=0 \end{gathered}[/tex]Take x as a common
[tex]\begin{gathered} x(\frac{x^3}{x}-\frac{4x}{x})=\frac{0}{x} \\ x(x^2-4)=0 \end{gathered}[/tex]Equate x by 0 and x^2 - 4 by 0 to find the values of x
[tex]\begin{gathered} x=0 \\ x^2-4=0 \\ x^2-4+4=0+4 \\ x^2=4 \\ \sqrt[]{x^2}=\pm\sqrt[]{4} \\ x=\pm2 \end{gathered}[/tex]The values of x are 0, 2, -2
Let us check them
[tex](4\times0)^{\frac{1}{3}}-0=0-0=0\rightarrow True[/tex][tex](4\times-2)^{\frac{1}{3}}-(-2)=(-8)^{\frac{1}{3}}+2=-2+2=0\rightarrow True[/tex][tex](4\times2)^{\frac{1}{3}}-(2)=(8)^{\frac{1}{3}}-2=2-2=0\rightarrow True[/tex]The equation has 3 possible solutions ----- 3rd answer
The last answer is -2, 0, 2 ------- 1st answer