The value of ${\int }_0^1{\left(2x^3-3x^2-x+1\right)}^{\frac{1}{3}}dx$ is equal to:

$I=\underset{0}{\overset{1}{\int }}{\left(2x^3-3x^2-x+1\right)}^{\frac{1}{3}}dx$

Using $\underset{0}{\overset{2a}{\int }}f\left(x\right)dx=0$ when$f\left(2a-x\right)=-f\left(x\right)$

Now, finding $f\left(1-x\right)$ we get,

$\Rightarrow I=\underset{0}{\overset{1}{\int }}{\left(2{\left(1-x\right)}^3-3{\left(1-x\right)}^2-\left(1-x\right)+1\right)}^{\frac{1}{3}}dx$

$\Rightarrow I=\underset{0}{\overset{1}{\int }}{\left(2\left(1-x^3-3x+3x^2\right)-3{\left(1+x^2-2x\right)}^2-\left(1-x\right)+1\right)}^{\frac{1}{3}}dx$

$\Rightarrow I=\underset{0}{\overset{1}{\int }}{\left(2-2x^3-6x+6x^2-3-3x^2+6x-1+x+1\right)}^{\frac{1}{3}}dx$

$\Rightarrow I=\underset{0}{\overset{1}{\int }}{\left(-2x^3+3x^2+x-1\right)}^{\frac{1}{3}}dx$

$\Rightarrow I=-I$

$\Rightarrow f\left(1-x\right)=-f\left(x\right)$

$\therefore I=0$