I am trying to show that $\int_{0}^{2}x^{3}\sqrt{8-x^3} dx=\frac{16\pi}{9 \sqrt{3}}$ using Beta function, i.e, $\beta(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}=\int_{0}^{1}t^{m-1}(1-t)^{n-1} dt$.
The attempt I made is this, I substitute that $x=2 t^{\frac{1}{3}}$, then $dx=\frac23 t^{\frac{-2}{3}} dt$, when I substitute this in the given integral, I get that $$\int_{0}^{2}x^{3}\sqrt{8-x^3} dx=32 \frac{\sqrt{2}}{3}\int_{0}^{1}t^{\frac{1}{3}}(1-t)^{\frac{1}{2}}dt$$ $$\Rightarrow ~ =32 \frac{\sqrt{2}}{3}\beta(\frac{4}{3},\frac{3}{2})$$ $$\Rightarrow ~ =32 \frac{\sqrt{2}}{3}\frac{\Gamma(\frac{4}{3})\Gamma(\frac{3}{2})}{\Gamma(\frac{17}{6})}$$ After simplication and using the properties of Gamma function, I get $$\Rightarrow ~ =8 \sqrt{2}\sqrt{\pi}\frac{8}{11}\frac{5}{6}\frac{\Gamma(\frac{1}{3})}{\Gamma(\frac{5}{6})}$$ I don't know how to can I get the desired result? Please help in detail if anyone can!