jmc

Cubing the given equation yields $$1=(1-x^3)+3\sqrt[3]{(1-x^3)(1+x^3)}\left(\sqrt[3]{1-x^3}+\sqrt[3]{1+x^3}\right)+(1+x^3)=2+3\sqrt[3]{1-x^6}.$$Then $\frac{-1}{3}=\sqrt[3]{1-x^6},$ so $\frac{-1}{27}=1-x^6$ and $x^6=\boxed{\frac{28}{27}}.$