How can one show that $\sqrt[3]{2}$ is not a root of a quadratic with rational coefficients?
It is clear that if $\sqrt[3]{2}$ is the root of such a quadratic, then it is also the root of a quadratic with integer coefficients. This reduces the problem a bit, but I don't see where one can go from here.
On
Consider the ring of polynomials $\def\Q{\mathbb{Q}}\Q[X]$ and the subset $$ I=\{f\in\Q[X]:f(\sqrt[3]{2})=0\} $$ I claim there exists a unique monic polynomial $p(X)\in I$ such that every polynomial $f\in I$ is of the form $f(X)=p(X)q(X)$ for some $q\in\Q[X]$.
In order to prove this we only need to know the division algorithm in $\Q[X]$: given $A(X)$ and $B(X)$ in $\Q[X]$, with $B\ne0$, there exist $Q(X)$ and $R(X)$ such that $$ A(X)=B(X)Q(X)+R(X)\text{ and the degree of $R$ is less than the degree of $B$}. $$ (we consider the degree of the zero polynomial to be $-\infty$).
In order to prove the claim about $I$, we just take a nonzero monic polynomial $p$ of minimum degree in $I$. Such a polynomial exists, because $X^3-2\in I$. Now, if $f\in I$, we can write $$ f(X)=p(X)q(X)+R(X) $$ where the degree of $R$ is less than the degree of $p$. Since $$ 0=f(\sqrt[3]{2})=p(\sqrt[3]{2})q(\sqrt[3]{2})+R(\sqrt[3]{2})=0\cdot q(\sqrt[3]{2})+R(\sqrt[3]{2}) $$ we conclude that $R(\sqrt[3]{2})=0$, so $R\in I$. By the minimality of the degree of $p$, the only possibility is that $R=0$. The uniqueness of $p$ is almost obvious: if $p_1$ is another nonzero monic polynomial in $I$, then $p-p_1\in I$ would have degree less than both, so it must be zero.
Now $p$ is irreducible: if $p=p_1p_2$, where $p_1$ and $p_2$ can be chosen monic, $$ 0=p(\sqrt[3]{2})=p_1(\sqrt[3]{2})p_2(\sqrt[3]{2}) $$ so at least one of the two factors must by in $I$ and, again, by the minimality of the degree of $p$ one of the two factors is $p$.
Now assume $p$ has degree less than $3$; then $$ X^3-2=p(X)q(X) $$ for some $q\in\Q[X]$. This is impossible, because $X^3-2$ is irreducible by Eisenstein's criterion. In particular $\sqrt[3]{2}$ is irrational.
This is standard theory. The set $I$, that can be defined for any algebraic number (that is, one that is root of a polynomial with rational coefficients), is an ideal of $\Q[X]$. The argument above can be applied to any ideal of $\Q[X]$ to show it is generated by a polynomial (we say that $\Q[X]$ is a principal ideal domain).
In the case of $I=\{f\in\Q[X]:f(r)=0\}$, where $r$ is an algebraic number, this ideal is also maximal and the generator (the unique nonzero monic polynomial of minimal degree) is called the minimal polynomial of $r$. Such a minimal polynomial is irreducible. Actually, it is not difficult to show that the minimal polynomial of $r$ can be characterized as the unique irreducible monic polynomial having $r$ among its roots.
On
I'll take it for granted that $2^{1/3}$ is irrational. (The proof is essentially the same as for the irrationality of $2^{1/2}$.)
Now suppose $2^{1/3}$ is the root of a quadratic polynomial with rational coefficients, $x^2+bx+c=0$. Then we can write
$$2^{2/3}=-b2^{1/3}-c$$
Multiplying by $2^{1/3}$ gives
$$\begin{align} 2&=-b2^{2/3}-c2^{1/3}\\ &=-b(-b2^{1/3}-c)-c2^{1/3}\\ &=(b^2-c)2^{1/3}+bc \end{align}$$
Since $2^{1/3}$ is irrational, we must have $c=b^2$, which leaves $2=bc=b^3$, so that $b=2^{1/3}$, which contradicts the assumption that $b$ is rational.
On
Below is a proof using only the fundamental theorem of algebra and polynomial long division (Euclid's algorithm), without using minimal polynomials or irreducibility criteria:
Let $p(x)$ be a rational-coefficient polynomial for which $\sqrt[3]{2}$ is a root.
Put $q(x)=x^3-2$ and consider $g(x)= \mbox{gcd}(p(x), q(x))$, obtained via Euclid's algorithm.
$g(x)$ has rational coefficients since $p(x)$ and $q(x)$ do, and $\deg(g(x))\le \deg(p(x))$. $p(x)$ and $q(x)$ share a root, so $g(x)$ has degree at least $1$. Denote the constant term of $g(x)$ by $a_0$.
The constant term of a polynomial is, up to a sign, the product of its roots and its leading coefficient*. $g(x)$ divides $q(x)$ so its roots are a subset of the roots of $q(x)$, which are the complex numbers $\omega_1, \omega_2, \omega_3$ each with modulus $\sqrt[3]{2}$. So taking modulus, we have $|a_0|= |a_n| \left(\sqrt[3]{2}\right)^n$ where $n$ is the degree of $g(x)$, and $a_n$ the leading coefficient of $g(x)$. In particular we get that $\left(\sqrt[3]{2}\right)^n$ is rational.
$\sqrt[3]{2}$ is not rational, so $n\neq 1$. And $\left(\sqrt[3]{2}\right)^2$ is not rational, so $n\neq 2$. So $n\ge 3$.
Since $n=\deg (g(x)) \le \deg (p(x))$ we obtain that $p(x)$ is of degree at least $3$.
*Proof: Expand $a_nx^n+...+a_1 x + a_0 = a_n(x-r_1)\cdots(x-r_n)$ and compare constant terms: $a_0 = a_n(-1)^nr_1\cdots r_n$.
It's already the root of some obvious cubic over ${\mathbb Z}$; that cubic is irreducible (why?), so it is the minimum polynomial of $\sqrt[3]{2}$; therefore no polynomial over ${\mathbb Q}$ of degree smaller than 3 has $\sqrt[3]{2}$ as root.