If $(a-b^2)b>0$, then $\sqrt[3]{a+\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}{3b}}}+\sqrt[3]{a-\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}{3b}}}$ is rational From Hardy´s "A course of pure mathematics" 10th edition, problem 31 miscellaneous problems of chapter I.

If $(a-b^2)b>0$, then
$$
\sqrt[3]{a+\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}{3b}}}+\sqrt[3]{a-\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}{3b}}}
$$
is rational.

"Note from the book":

Each of numbers under a cube root is of the form
$$
\left( \alpha + \beta\sqrt{\frac{a-b^3}{3b}} \right)^3
$$
where $\alpha$ and $\beta$ are rational.


Hint . Put $$a=y^3$$
We get :
$$b\Bigg(\sqrt[3]{x^3+\Big(3+\frac{x^3}{3}\Big)\sqrt{\frac{x^3-1}{3}}}+\sqrt[3]{x^3-\Big(3+\frac{x^3}{3}\Big)\sqrt{\frac{x^3-1}{3}}}\Bigg)\quad (1)$$
Where $x=\frac{y}{b}$
Now if $b$ is rational the expression in front of $b$ must be a rational but I don't see how ...
In fact if $x=1$ the expression $(1)$ becomes :
$$2y$$
Or :
$$2\sqrt[3]{a}$$
If we follow the comment of  @Paramanand Singh there is a nice inverse function to one of the summand in the bracket of the expression $(1)$ see here
