If a real number can be expressed in terms of complex solutions of cubic equations, can it be expressed in terms of real solutions of cubic equations?

For the constructible numbers, any quadratic extension is a radical extension, and there must be a first non-real field, which is gotten by adjoining $\sqrt a$ for some real $a<0$; then we can just write $\sqrt a=i\sqrt{|a|}$. Thus, after repeatedly applying the formula for complex square roots, performing the required complex arithmetic, and finally discarding the $i$ component, any constructible $\alpha\in\mathbb R$ can be written in terms of real square roots.

Similarly, for the cubic-constructible numbers, there must be a first non-real field, which is gotten by adjoining $\beta=\omega\sqrt[3]{a+\sqrt b\,}+\omega^2\sqrt[3]{a-\sqrt b\,}$ where $b>0$. This has the form $c+di$ where $c$ and $d$ are real-cubic-constructible:

$$\beta=\frac{-1+i\sqrt3}{2}\sqrt[3]{a+\sqrt b\,}+\frac{-1-i\sqrt3}{2}\sqrt[3]{a-\sqrt b\,}$$

$$=-\frac12\left(\sqrt[3]{a+\sqrt b\,}+\sqrt[3]{a-\sqrt b\,}\right)+\frac{\sqrt3}{2}\left(\sqrt[3]{a+\sqrt b\,}-\sqrt[3]{a-\sqrt b\,}\right)i.$$

Clearly, complex numbers of this form are closed under field operations. They're also closed under cube roots:

$$(x+yi)^3=a+bi$$

$$x^3-3xy^2=a,\quad3x^2y-y^3=b$$

$$3xy^2=x^3-a,\quad(3x^2-y^2)y=b$$

$$\big(9x^3-(x^3-a)\big)y=3bx$$

$$(8x^3+a)^2y^2=9b^2x^2$$

$$(8x^3+a)^2(x^3-a)-27b^2x^3=0.$$

This equation is $9$th degree in $x$, but it's $3$rd degree in $x^3$; furthermore, evaluating at $x^3\to-\infty,\;x^3=-a/8,\;x^3=a,\;x^3\to+\infty$ shows three sign changes, so we can solve for three different values of $x^3$. Taking a real cube root gives $x$, and the middle equation gives $y$ in terms of $x$. So we get three cube roots of $a+bi$. (This applies to the general case $ab\neq0$, but the special cases are easy to handle.)