<s>[INST] Is $\sqrt[3]{7+5 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ a rational number? 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.
 [/INST] 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 $\sqrt[3]{7+5\sqrt2}+\sqrt[3]{20-14\sqrt2}$.
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