<s>[INST] 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.
 [/INST] 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.
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