Prove that $5^{1/3}+7^{1/2}$ is irrational

Suppose that our sum is equal to the rational number $r$. Then $5^{1/3}=r-\sqrt{7}$. Cubing both sides we obtain $$5=r^3-3r^2\sqrt{7}+21r-7\sqrt{7},$$ and therefore $$r^3+21r-5-(3r^2+7)\sqrt{7}=0.$$ Since $\sqrt{7}$ is irrational, we must have $3r^2+7=0$. This is impossible.

Remark: For completeness we sketch a proof of the irrationality of $\sqrt{7}$. It is much like one of the standard proofs of the irrationality of $\sqrt{2}$. Suppose to the contrary that there are integers $p$ and $q$, with $q\ne 0$, such that $p^2/q^2=7$. Without loss of generality we may assume $p$ and $q$ are relatively prime.

Then $p^2=7q^2$. Since $7$ is prime and divides $p^2$, it follows that $7$ must divide $p$. Let $p=7t$. Substituting and cancelling, we get $7t^2=q^2$. Thus $7$ divides $q$, contradicting the fact that $p$ and $q$ are relatively prime.

Let $t=5^{1/3}+7^{1/2}$. Then $t$ is a root of $x^6-21 x^4-10 x^3+147 x^2-210 x-318=0$. (*)

The rational root theorem tell us that either $t$ is irrational or $t$ is an integer dividing $318$.

Since $ 1 < 5^{1/3} < 2$ and $ 2 < 7^{1/2} < 3$, we have $ 3 < t < 5$, and so if $t$ is an integer, it must be $4$. But $4$ does not divide $318$.

Therefore, $t$ is irrational.

(*) This polynomial can be found manually (if you don't drown in the algebra) or you can ask WA. This is not cheating (well, not crucially) because the nice argument is still to come. It's an opportunity to learn the rational root theorem, if you haven't seen it.

Another solution is to look at $K = \mathbb{Q}(\zeta_3, \sqrt[3]{5}, \sqrt[2]{7})$ (where $\zeta_3$ is primitive cube root of unity) which is Galois over $\mathbb{Q}$. It is easy to see that $(K : \mathbb{Q}) = 6 \times 2 = 12$ and the automorphism in $\text{Gal}(K|\mathbb{Q})$ are easily described: they are of the form $\sigma_{i,j,\pm} : \{\zeta_3 \mapsto \zeta_3^i\} \times \{\sqrt[3]{5} \mapsto \sqrt[3]{5} \zeta_3^j\} \times \{\sqrt[2]{7} \mapsto \pm \sqrt[2]{7}\}$ where $i \in \{1, 2\}, j \in \{0, 1, 2\}$. It is easy to check that one of the automorphism $\sigma_{1,2,+}$ does not fix $\alpha = \sqrt[3]{5} + \sqrt[2]{7}$. Hence, $\alpha \not\in \mathbb{Q}$.