Using a cubic equation to prove a radical

Question

By considering $x^3-15x-4=0$, prove that

$$ \sqrt[\leftroot{-2}\uproot{2}3]{2 + \sqrt{-121}} + \sqrt[\leftroot{-2}\uproot{2}3]{2 - \sqrt{-121}} = 4$$

Proving this in other ways may be simple (?), but I'm very interested to know how this equation suggested could be used in this example, in particular we have that $\sqrt{-121}=11i$, so I don't know where the $-15x$ comes from. Cubing both sides won't work either... is there a known method or theorem for this?

Answer 1

Hint:   let $a=\sqrt[3]{2 + \sqrt{-121}}$ and $b = \sqrt[3]{2 - \sqrt{-121}}$, then $a^3+b^3=4$ and $ab = \sqrt[3]{125}=5$, so:

$$ \underbrace{\color{blue}{(a+b)^3}}_{x^3}=a^3+b^3+3ab(a+b) = \underbrace{\color{blue}{4 + 15(a+b)}}_{4+15x} $$

It follows that $a+b$ is a root of $x^3 - 15 x - 4 =0$. The cubic has only one positive root, which can be easily found using the rational root theorem.

---

[ EDIT ]   The reason to look for the positive real root here is that $\,b = \bar a\,$, so $a+b = 2 \operatorname{Re}(a)$. But $a^3= 2 + 11i$ is quite obviously in the first quadrant, so the principal value of its cube root must be in the first quadrant, too, in other words $\operatorname{Re}(a) \gt 0$.

Answer 2

Since$$(2+i)^3=2^3+3\times2^2i+3\times2i^2+i^3=2+11i$$and $(2-i)^3=2-11i$, it is natural to take$$\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=2+i+2-i=4.$$

Answer 3

Note that $$x^3-15x-4=(x-4)(x^2+4x+1)$$

Thus $x=4$ is the only positive root of $$x^3-15x-4$$

We show, $$x=\sqrt[\leftroot{-2}\uproot{2}3]{2 + \sqrt{-121}} + \sqrt[\leftroot{-2}\uproot{2}3]{2 - \sqrt{-121}} $$

Satisfies the equation $$x^3-15x-4 =0$$

Upon using the formula $$(a+b)^3 = a^3+3a^2b+3ab^2+b^3=( a^3 + b^3)+ 3ab(a+b) $$

with $$a=\sqrt[\leftroot{-2}\uproot{2}3]{2 + \sqrt{-121}} $$

and $$b= \sqrt[\leftroot{-2}\uproot{2}3]{2 - \sqrt{-121}} $$ we get the desired result.