If $\sqrt[3]{a} + \sqrt[3]{b}$ is rational then prove $\sqrt[3]{a}$ and $\sqrt[3]{b}$ are rational

As you have proved, $p^2+q^2\in\Bbb Q$ and $pq\in \Bbb Q$, so $$p-q=\frac{a-b}{p^2+pq+q^2}\in \Bbb Q.$$ Combining this with $p+q\in \Bbb Q$, we are done.

Statement for the general case (for any $n>1$ in place of $3$)

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 we show that $\sqrt[n]{a}$ is the unique common root of $p$ and $q$. Let $z$ be a common root of $p$ and $q$. Then $z=\sqrt[n]{a}\omega_1$ and $s-z=\sqrt[n]{b}\omega_2$ for some $n^{th}$ roots of unity $\omega_1,\omega_2$. Then $$\sqrt[n]{a}+\sqrt[n]{b}=\sqrt[n]{a}\omega_1+\sqrt[n]{b}\omega_2\tag{1}$$ Since $|\omega_1|=|\omega_2|=1$ therefore $|\mathrm{Re}(\omega_i)|\leq1$ with equality if and only if $\omega_i=\pm1$ for $i\in\{1,2\}$. Now comparing real parts in the equation $(1)$ we can easily conclude that $\omega_1=\omega_2=1$ implying that $\sqrt[n]{a}$ is the unique common root.

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.