We have, $f(x)=x(x-1)^2$ in $[0,1]$.
(i) Since, $\mathrm{f}(\mathrm{x})=\mathrm{x}(\mathrm{x}-1)^2$ is a polynomial function. So, it is continuous in $[0,1]$.
(ii) Now, $f^{\prime}(\mathrm{x})=\mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}-1)^2+(\mathrm{x}-1)^2 \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}$ $=3 x^2-4 x+1$ which exists in $(0,1)$ So, $\mathrm{f}(\mathrm{x})$ is differentiable in $(0,1)$.
(iii) Now, $f(0)=0$ and $f(1)=0 \Rightarrow f(0)=f(1)$
f satisfies the above conditions of Rolle's theorem.
Hence, by Rolle's theorem $\exists \mathrm{c} \in(0,1)$ such that
$$
\begin{aligned}
\mathrm{f}^{\prime}(c) &=0 \\
\Rightarrow \quad \quad 3 c^2-4 c+1 &=0 \\
\Rightarrow \quad(3 c-1)(c-1) &=0 \\
\Rightarrow \quad \mathrm{c}=\frac{1}{3}, 1 \Rightarrow \frac{1}{3} \in(0,1)
\end{aligned}
$$
Thus, we see that there exists a real number $\mathrm{c}$ in the open interval $(0,1)$.
Hence, Rolle's theorem has been verified.