The maximum value of

Let $f(x)=[x(x-1)+1]^{\frac{1}{3}}$.

$\therefore f^{\prime}(x)=\frac{2 x-1}{3[x(x-1)+1]^{\frac{2}{3}}}$

Now, $f^{\prime}(x)=0 \Rightarrow x=\frac{1}{2}$

Then, we evaluate the value of $f$ at critical point $x=\frac{1}{2}$ and at the end points of the interval $[0,1]\{$ i.e., at $x=0$ and $x=1\}$.

$f(0)=[0(0-1)+1]^{\frac{1}{3}}=1$

$f(1)=[1(1-1)+1]^{\frac{1}{3}}=1$

$f\left(\frac{1}{2}\right)=\left[\frac{1}{2}\left(\frac{-1}{2}\right)+1\right]^{\frac{1}{3}}=\left(\frac{3}{4}\right)^{\frac{1}{3}}$

Hence, we can conclude that the maximum value of f in the interval [0, 1] is 1.

The correct answer is C.