<s>[INST] 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)
 [/INST] Assume  that $$ \sqrt[3]{5} + \sqrt{2}=r$$ where r is a rational number.
We have $$ \sqrt[3]{5} =r-\sqrt{2}$$
Raise to the third power to get $$5=r^3-3r^2 \sqrt 2 +6r - 2\sqrt 2 $$
Solving for $\sqrt 2$ we get $$ \sqrt 2 =  \frac {5-r^3-6r}{-3r^2-2}$$ 
The $RHS$  is a rational number which is impossible.
</s>