Is $\sqrt[3]{7+5 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ a rational number?

Question

Is there a way to show that

$$\alpha=\sqrt[3]{7+5 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$$

is a rational number?

I found $\alpha=3$ from doing simplifications. But, I would like to known a different approach.

Answer 1

Anay's idea of cubing both sides looks promising, but there is a little subtlety in the method. Let's explore.

Start with

$\alpha=\sqrt[3]{7+5\sqrt2}+\sqrt[3]{20-14\sqrt2}$

Cube with the Binomial Theorem:

$\alpha^3=\color{blue}{7+5\sqrt2}+\color{brown}{3(\sqrt[3]{7+5\sqrt2})^2\sqrt[3]{20-14\sqrt2}+3\sqrt[3]{7+5\sqrt2}(\sqrt[3]{20-14\sqrt2})^2}+\color{blue}{20-14\sqrt2}$

Combine the terms that are the same color in the above expression:

$\alpha^3=\color{blue}{27-9\sqrt2}+\color{brown}{3\sqrt[3]{7+5\sqrt2}\sqrt[3]{20-14\sqrt2}(\sqrt[3]{7+5\sqrt2}+\sqrt[3]{20-14\sqrt2})}$

$\alpha^3=27-9\sqrt2+3\sqrt[3]{2\sqrt2}\alpha$

Render $\sqrt[3]{2\sqrt2}=\sqrt2$:

$\alpha^3-3\sqrt2\alpha-27+9\sqrt2=0$

Now suppose that $\alpha$ is rational in this final equation. Then the left side is rational only if $\sqrt2$ has a zero overall coefficient, meaning $3\alpha-9=0$ or $\alpha=3$ is the only possible rational root. So, we try this value and find that the equation holds. Thus $\alpha=3$ checks out.

We have the possibility that even though $\alpha=3$ is a root of the cubic equation, it may not be the original sum of the cube roots with which we started. To check that, divide the cubic by $\alpha-3$ to get a quadratic equation whose remaining real roots are other possibilities for $\alpha$. This gives the factorization

$(\alpha-3)(\alpha^2+3\alpha+9-3\sqrt2)=0$

The discriminant of the quadratic factor is $-27+12\sqrt2=-\sqrt{729}+\sqrt{288}<0$. Thereby $\alpha=3$ is the only real root and must be the actual sum $\sqrt[3]{7+5\sqrt2}+\sqrt[3]{20-14\sqrt2}$.

Answer 2

Note $\sqrt[3]{20-14 \sqrt{2}}=\sqrt2 \cdot \sqrt[3]{5 \sqrt{2}-7}$ and assume $\sqrt[3]{5\sqrt{2}\pm7}=p\sqrt2\pm q$. Then

$$\alpha =(p\sqrt2+q)+\sqrt2(p\sqrt2-q)=(q+2p)+(p-q)\sqrt2 $$

For $a$ to be rational requires $p=q=\frac{\sqrt[3]{5 \sqrt{2}+7}}{\sqrt2+1} = \sqrt[3]{\frac{ 5 \sqrt{2}+7}{(\sqrt2+1)^3 }}=1$ to be rational, which is indeed.