Cubic equation

$\sqrt[3]{x^3-3x^2}+1=x\\\sqrt[3]{x^3-3x^2}=x-1\\x^3-3x^2=x^3-3x^2+3x-1\\3x-1=0\rightarrow x=\frac{1}{3}$

We have $\sqrt[3]{x^3 - 3x^2} + 1 = x.$ After raising both sides of the equation to the third power it follows that $x^3 - 3x^2 = \left(x-1\right)^3 = x^3 - 3x^2 + 3x - 1.$ Now we cancel the like terms and we are left with $x = \frac{1}{3}.$

This tells us that if $x$ is a solution to the original equation it must be equal to $\dfrac{1}{3}$ . Substituting $x = \frac{1}{3}$ to the original equation yields $\begin{aligned} \sqrt[3]{\left(\dfrac{1}{3}\right)^3 - 3\left(\dfrac{1}{3}\right)^2} + 1 &= \dfrac{1}{3} \\ \sqrt[3]{\dfrac{1}{27} - \dfrac{9}{27}} + 1 &= \dfrac{1}{3} \\ \sqrt[3]{-\dfrac{8}{27}} + 1 &= \dfrac{1}{3} \\ -\dfrac{2}{3} + 1 &= \dfrac{1}{3} \\ \dfrac{1}{3} &= \dfrac{1}{3}. \\ \end{aligned}$

Which tells us that $x = \frac{1}{3}$ is indeed a solution to the original equation.

Hence the solution is $\boxed{\dfrac{1}{3}}$ .