Let $a$ and $b$ be two positive rational numbers such that $\sqrt[3] {a} + \sqrt[3]{b}$ is also a rational number. Prove that $\sqrt[3]{a}$ and $\sqrt[3] {b}$ themselves are rational numbers.
My first response was, isn't is obvious? but then I tried..
if $a=b$ then we prove it easily . So suppose $a\ne b$.
Let $s=\sqrt[3]{a} + \sqrt[3]{b}$.
then $s^3 = a + b + 3 s \sqrt[3]{a} \sqrt[3]{b}$
then the product $p=\sqrt[3]{a} \sqrt[3]{b}$ is a rational number.
But what should I do next ?
Statement for the general case
Let $a,b$ be two positive rational numbers such that $\sqrt[n]{a}+\sqrt[n]{b}$ is a rational number for some natural number $n>1$. Then both $\sqrt[n]{a}$, $\sqrt[n]{b}$ are rational numbers.
Proof
Let $s=\sqrt[n]{a}+\sqrt[n]{b}\in\mathbb{Q}$. Then $\sqrt[n]{a}$ is a root of the polynomial $p(X)=X^n-a$ (If $\sqrt[n]{a}$ is not rational then $p$ is in fact the minimal polynomial). Again $\sqrt[n]{a}$ is also a root of $q(X)=(s-X)^n-b$. Now try to prove that $\sqrt[n]{a}$ is the unique common root of $p$ and $q$. Then $$\gcd(p(X),q(X))=X-\sqrt[n]{a}$$ But $p(X),q(X)\in\mathbb{Q}[X]$ implies $X-\sqrt[n]{a}$ is in $\mathbb{Q}[X]$ then $\sqrt[n]{a}\in\mathbb{Q}$. Similarly one can prove that $\sqrt[n]{b}$ is also rational.
From your work, we can write $$\frac{p}{\sqrt[3]{a}} + \sqrt[3]{a} = s,$$ so $$\sqrt[3]{a} = \frac{s\pm \sqrt{s^2-4p}}{2}.$$
Cube both sides and solve for $\sqrt{s^2-4p}$ (noting $s^2-p \neq 0$) to show it's rational, which then shows $\sqrt[3]{a}$ is rational.
we have $\sqrt[3]{a}+\sqrt[3]{b}$ is a rational number and so is their product.(which you found)
Now, squaring $\sqrt[3]{a}+\sqrt[3]{b}$ we can get, $\sqrt[2/3]{b}+\sqrt[2/3]{a}$ is also rational(using the fact that $\sqrt[3]{ab} \in \mathbb Q$ Similarly,squaring $\sqrt[2/3]{b}+\sqrt[2/3]{a}$
we can get, $\sqrt[4/3]{b}+\sqrt[4/3]{a}=a*\sqrt[3]{a}+b*\sqrt[3]{b} \in \mathbb Q$ Now,multiply $\sqrt[3]{a}+\sqrt[3]{b}$ by a and subtract it from $a*\sqrt[3]{a}+b*\sqrt[3]{b}$ to get the desired result.
