Given \(f(x)=x^3-8\)

Set \(x^3-8\) equal to \(0\)

Solve for \(x\).

Add \(8\) to both sides of the equation.

\(x^3=8\)

Subtract \(8\) from both sides of the equation.

\(x^3−8=0\)

Factor the left side of the equation

\((x−2)(x^2+2x+4)=0\)

If any individual factor on the left side of the equation is equal to \(0\), the entire expression will be equal to \(0\).

\(x−2=0\)

\(x^2+2x+4=0\)

Set \(x−2\) equal to \(0\) and solve for \(x\).

\(x=2\)

Set \(x^2+2x+4\) equal to \(0\) and solve for \(x\).

\(x=−1+i \sqrt3,−1−i \sqrt3\)

The final solution is all the values that make \((x−2)(x^2+2x+4)=0\) true

Answer: \(x=2,−1+i \sqrt3,−1−i \sqrt3\)