How to show that $\sqrt[3]{-1+\sqrt{-7}}+\sqrt[3]{-1-\sqrt{-7}}$ is a real number at a time before the invention of complex numbers I have read this PDF from ocw.mit.edu about complex numbers. There is one interesting question: Imagine yourself at the time, when complex numbers had to be invented yet. How to show that $$\sqrt[3]{-1+\sqrt{-7}}+\sqrt[3]{-1-\sqrt{-7}}$$ is a real number.
 A: The polynomial $x^3-6x+2=0$ has three real roots, which is easy to prove. By Cardano's method, one of the roots is just the above expression. So it must be real. If we do not believe the existence of complex numbers though, the expression is not real, because it does not exist.
A: Let $$r = (-1 + \sqrt{-7})^{1/3} + (-1 - \sqrt{-7})^{1/3},$$ where by $\sqrt{-7}$ we mean some number $x$ whose square is $-7$; i.e., $x^2 = -7$.  Recall the identity $$(a^{1/3} + b^{1/3})^3 = a + b + 3(ab)^{1/3}(a^{1/3} + b^{1/3}).$$  Consequently, $$r^3 = -2 + 3((-1)^2 + 7)^{1/3}r = 6r - 2.$$  Now consider the trigonometric identity $$\begin{align*} \cos 3\theta &= \cos \theta \cos 2\theta - \sin \theta \sin 2\theta \\ &= \cos^3 \theta - \cos \theta \sin^2 \theta - 2 \sin^2 \theta \cos \theta \\ &= \cos^3 \theta - 3 \cos\theta \sin^2 \theta \\ &= \cos^3 \theta - 3 \cos\theta (1 - \cos^2 \theta) \\ &= 4 \cos^3 \theta - 3 \cos \theta. \end{align*}$$  This suggests the choice $r = 2 \sqrt{2} \cos \theta$ gives $$\begin{align*} r^3 - 6r + 2 &= 16 \sqrt{2} \cos^3 \theta - 12 \sqrt{2} \cos \theta + 2 \\ &= 4 \sqrt{2} (4 \cos^3 \theta - 3 \cos \theta) + 2 \\ &= 4 \sqrt{2} \cos 3\theta + 2 = 0. \end{align*}$$  Consequently, $$\theta = \frac{1}{3}\cos^{-1}\left( - \frac{1}{2 \sqrt{2}} \right),$$ hence $$r \in 2 \sqrt{2} \cos \left( \frac{1}{3} \cos^{-1} \left( - \frac{1}{2\sqrt{2}} \right) + \frac{2\pi k}{3}\right), \quad k = 0, 1, 2,$$ all of which are real values.  Of course, this result presupposes that the rules that allow us to work with real numbers also work for numbers like $\sqrt{-7}$.  To be completely rigorous, in my opinion, requires an axiomatic treatment of the complex numbers.  The above really only demonstrates that $r$ is real if the rules we use for arithmetic are extended to such values.
A: I think it's a stretch to claim that this result can be proven without complex numbers, especially because of the presence of $\sqrt{-7}$. Anyway, here's a way to look at it. Let $A = \sqrt[3]{-1 + \sqrt{-7}}$ and $B = \sqrt[3]{-1 - \sqrt{-7}}$. If $t = A + B$, then $$t^3 = A^3 + B^3 + 3AB(A + B) = (-1 + \sqrt{-7}) + (-1 - \sqrt{-7}) + 3t\sqrt[3]{1 - (-7)} = -2 + 6t$$ So $t$ is a root of the cubic $x^3 - 6x + 2$. Let $f(x) = x^3 - 6x + 2$. Then $f(-3) = -7 < 0$, $f(0) = 2 > 0$, $f(1) = -3 < 0$, and $f(3) = 11 > 0$. Hence, by the intermediate value theorem, $f$ has three real roots. Therefore, $t$ is real.
