How to evaluate this integral - beta function? $$ \displaystyle\int\limits^{\cssId{upper-bound-mathjax}{1}}_{\cssId{lower-bound-mathjax}{0}} \left(1-\left(1-x^3\right)^\sqrt{2}\right)^\sqrt{3}x^2\,\cssId{int-var-mathjax}{\mathrm{d}x} $$
I have been taught Gamma function and Beta function recently. But I cannot evaluate this integral. I am unable to proceed. I need help with the approach. 
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}\bracks{1 - \pars{1 - x^{3}}^{\root{2}}}^{\root{3}}x^{2}\,\dd x}
\\[5mm] \stackrel{x^{3}\ \mapsto\ x}{=}\,\,\,&
{1 \over 3}\int_{0}^{1}
\bracks{1 - \pars{1 - x}^{\root{2}}}^{\root{3}}\dd x
\\[5mm]
\stackrel{1 - x\ \mapsto\ x}{=}\,\,\,&
{1 \over 3}\int_{0}^{1}
\bracks{1 - x^{\root{2}}}^{\root{3}}\dd x
\\[5mm] \,\,\,\stackrel{\pars{1 - x}^{\root{2}}\ \mapsto\ x}{=}\,\,\,&
{\root{2} \over 6}\int_{0}^{1}t^{\root{3}}\pars{1 - t}^{\root{2}/2 - 1}
\,\dd t
\\[5mm] = &\
\bbx{{\root{2} \over 6}\,\mrm{B}\pars{\root{3} + 1,{\root{2} \over 2}}}
\approx 0.1546
\end{align}

$\ds{\mrm{B}}$ is the
  Beta Function.

A: Hint: Try the substitution $u=(1-x^3)^\sqrt{2}$.
