Prove that $\sqrt[3]{5} + \sqrt{2}$ is irrational I tried with both squaring and cubing the statement, it got messy, here's my  latest attempt: 
Assume for the sake of contradiction: $\sqrt[3]{5} + \sqrt{2}$ is rational
$\sqrt[3]{5} + \sqrt{2}$ = $\frac{a}{b}$ $a,b$ are odd integers $> 0$ and $ b\neq 0$ 
${(\sqrt[3]{5} + \sqrt{2})}^3$ = $\frac{a^3}{b^3}$
by multiplying by $b^3$: 
${(\sqrt[3]{5} + \sqrt{2})}^3 \times b^3 $ = ${a^3}$
so: $a^3$ is divisible by ${(\sqrt[3]{5} + \sqrt{2})}^3$ which means $a$ is divisible by ${(\sqrt[3]{5} + \sqrt{2})}$
doing the same thing with $b$ i found : 
$\frac{a^3}{{(\sqrt[3]{5} + \sqrt{2})}^3}   $ = ${b^3}$
so: $b^3$ is divisible by ${(\sqrt[3]{5} + \sqrt{2})}^3$ which means $b$ is divisible by ${(\sqrt[3]{5} + \sqrt{2})}$  (wrong)
${(\sqrt[3]{5} + \sqrt{2})}$ is a common divisor for both $a$ & $b$ which is a contradiction, thus $\sqrt[3]{5} + \sqrt{2}$ is irrational. (wrong)

You can't do divisibility in irational and rational numbers. When you are operating with divisibility you have to have an integers. It is a relation defined on integer numbers.

Suppose it is rational, then exist rational number $q$ such that $$\sqrt[3]{5} + \sqrt{2}= q$$ so $$ 5 = (q-\sqrt{2})^3 = q^3-3q^2\sqrt{2}+6q-2\sqrt{2}$$
So we have $$\sqrt{2}(\underbrace{3q^2+2}_{\in\mathbb{Q}}) = \underbrace{q^3+6q-5}_{\in\mathbb{Q}}$$
so $$\sqrt{2}=  \underbrace{q^3+6q-5\over 3q^2+2}_{\in\mathbb{Q}}$$
A contradiction.
