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

Question

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.

Answer 1

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

Answer 2

No proof, but making the start. From the hint we have $$\left( \alpha + \beta\sqrt{\frac{a-b^3}{3b}} \right)^3 = a+\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}{3b}} \\ =\frac{\beta \left(\beta^2(a-b^3)+9\alpha^2b\right)}{3b} \sqrt{\frac{a-b^3}{3b}} + \frac{\alpha\left(\beta^2(a-b^3) + \alpha^2b\right)}{b}$$ so by comparison we have the set of equations $${\alpha\left(\beta^2(a-b^3) + \alpha^2b\right)}=ab \tag{1}$$ $${\beta \left(\beta^2(a-b^3)+9\alpha^2b\right)} = {9b^3+a} \tag{2} \, .$$ Replacing $\beta^2(a-b^3)$ from (1) into (2) gives $$\beta=\frac{9b^3+a}{\frac{ab}{\alpha}+8\alpha^2 b} \, .$$ This shows that if $\alpha$ is rational, then so is $\beta$. Furthermore, if $(\alpha,\beta)$ is a solution to (1)+(2), then $(\alpha,-\beta)$ is a solution to (1)+(2) with $-\frac{9b^3+a}{3b}$ instead of $+$. So the sought after expressions for the cube roots are indeed "conjugate" to each other.

Now, plugging $\beta$ into (1) gives the equation $$\left(\frac{ab}{\alpha}+8\alpha^2b\right)^2\left(\frac{a}{\alpha}-\alpha^2\right)=\left(9b^3+a\right)^2 \left(\frac{a}{b}-b^2\right) \, .$$

Multiplying by $\alpha^3$ leads to a cubic polynomial in $\alpha^3$ for which rational solutions $\alpha=\frac{p}{q}$ are sought (gcd(p,q)=1). Again, inserting this expression into the polynomial and collecting by common denominator leads to the equation $$-64\,{b}^{3}{p}^{9}+48\,a{b}^{3}{p}^{6}{q}^{3}+ \left( 81\,{b}^{9}-63 \,a{b}^{6}-2\,{a}^{2}{b}^{3}-{a}^{3} \right) {q}^{6}{p}^{3}+{a}^{3}{b} ^{3}{q}^{9}=0 \, .$$ We can assume that $a,b$ are integers (otherwise multiply by a power of the common denominator of a and b), so reducing this equation mod $p^3$ gives $p|ab$ and mod $q^3$ gives $q|4b$.

to be continued.