$x^3-2$ over $\mathbb{R}$ and over $\mathbb{C}$

By [L1], the polynomial $f$ has a real root $r$. Since $r^3=2>0$, [L3] gives $r>0$. The identity $x^3-r^3=(x-r)(x^2+rx+r^2)$ therefore yields $x^3-2=(x-r)(x^2+rx+r^2)$ over $\mathbb{R}$.

If $x^2+rx+r^2=0$, then $(2x+r{)}^2=4x^2+4rx+r^2=-3r^2,$ which is impossible in $\mathbb{R}$ because the left-hand side is nonnegative while the right-hand side is negative. So the quadratic factor is irreducible over $\mathbb{R}$.

By [L2], let $\sqrt{3}>0$ be the real square root of $3$. Solving the quadratic factor from step 1.1 gives the two nonreal roots $\frac{-r\pm i\sqrt{3}r}{2}\mathrm{.}$ Substituting them back into the factorization from step 1.1 gives the displayed complete factorization over $\mathbb{C}$.

Thus $x^3-2$ has one real linear factor and one irreducible quadratic factor over $\mathbb{R}$, but it splits completely into three linear factors over $\mathbb{C}$.