Prove that you can't write $\sqrt[3]{4}$ like $a+b\sqrt[3]{2}$

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.

Let us write $x=2^{1/3}$. Then the statement to be shown false, which we will presume true to obtain a contradiction, may be written$$x^2=a+bx$$for some rational $a$ and $b$. Then$$2=x^3=xx^2=ax+bx^2=ab+(a+b^2)x.$$Since $x$ is irrational (the proof of which is very similar to the proof that $\surd2$ is irrational), it follows that $a+b^2=0$. From this we get $x^2=-b^2+bx$, or$$x^2-bx+b^2=0.$$But this quadratic equation has no real root (in $x$ or in $b$)—a contradiction.