# 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.

• You are right: if you don't assume the existence of complex numbers, this expression is meaningless. But this is exactly how complex numbers came to be invented. Jan 29, 2015 at 20:12
• fiddle with Cardano's method and see what equation of type $x^3 + p x + q = 0$ this expression solves. I'm not saying this really leaves complex numbers out of the picture... Jan 29, 2015 at 20:19
• @TonyK I didn't express myself clearly. I was imagining myself in the situation before the invention of complex numbers. At that time negative roots did not make sense. Then, how would I know that this expression above makes sense at all (i.e. the result is a real number)? Jan 29, 2015 at 20:23
• I don't see any misunderstanding between us! Cardano didn't think it made sense either, but it gave results, so he used it. It reminds me of the 20th-century quantum field theorists who discovered that they could reproduce experimental results by subtracting infinity from infinity to leave a finite quantity. (This last problem still hasn't been resolved, as far as I know.) Jan 29, 2015 at 22:46

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.
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.
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.