<s>[INST] Prove that you can't write $\sqrt[3]{4}$ like $a+b\sqrt[3]{2}$ 
Prove that you can't write $\sqrt[3]{4}$ like $a+b\sqrt[3]{2}$, with $a,b \in \Bbb Q$

How I can prove it? I tried elevating to cube but then?
$4 = a^3+ 3a^2b \sqrt[3]{2} + 3ab^2\sqrt[3]{4}+ 2b^3$
 [/INST] Assume for contradiction that $\sqrt[3] 4 = a + b \sqrt[3] 2$. We then cube both sides, to obtain $4 = a^3 + 3a^2b \sqrt[3] 2 + 3ab^2 \sqrt[3] 4 + 2b^3$. We have that $4, a^3, 2b^3$ are all rational, so we also have $3a^2b \sqrt[3] 2 + 3ab^2 \sqrt[3] 4 = q$ for some rational $q$. But $\sqrt[3] 4 = (\sqrt[3] 2)^2$, so we obtain a quadratic in $\sqrt[3] 2$ with rational coefficients, implying that $\sqrt[3] 2$ can be written as some $c \pm \sqrt d$. Clearly $d$ is nonzero since $\sqrt[3] 2$ is irrational, so cubing both sides again we obtain $c^3 \pm c^2 \sqrt d + cd \pm d\sqrt d = 2$, implying that $(c^2+d) \sqrt 2$ is rational. But $c^2+d$ is rational, and $\sqrt 2$ is not, yielding a contradiction.
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