# How to denest $\sqrt[3]{126i\sqrt{3}-55}$?

I was trying to solve the equation $x^3-2x^2-11x+12=0$ using Cardano's method, and I found myself with the following nested radical: $$\sqrt[3]{126i\sqrt{3}-55}$$

Is there any way to simplify this? I guess it has because I know from advance that this equation has nice solutions. Although, I cannot simplify it, even after researching on the subject. Most methods I tried take me to more nested radicals or more cubic equations.

• $\sqrt i$ isn't really defined... – Frank Nov 20 '16 at 23:49
• Ok, most probably this is multivalued, but I need at least one possible value. – J. C. Nov 20 '16 at 23:53
• Sure, I'll write an answer... – Frank Nov 20 '16 at 23:54

Hint:  let $\,a=\sqrt[3]{126i\sqrt{3}-55}\,$ and $\,b=\sqrt[3]{126i\sqrt{3}+55}\,$. Then:

$$a^3 - b^3 = -110$$ $$ab = \sqrt[3]{-126^2 \cdot 3 - 55^2} = \sqrt[3]{-50653} = -37$$

Writing $a^3-b^3 = (a-b)(a^2+ab+b^2) = (a-b)\big((a-b)^2 + 3ab)$ and letting $c=a-b$ gives:

$$c(c^2-3 \cdot 37) = -110$$

$$c^3 - 111 c + 110 = 0$$

Factoring out the obvious root $c=1$ leaves a quadratic which gives the other two roots $\{-11,10\}$.

For each $c$, the values $a,b$ can be obtained by solving the quadratic with integer coefficients that results from $a-b=c\,, \;ab=-37\,$.

• Thank you, I was able to get the solutions with your answer. Nevertheless, I am not satisfied yet: how could I denest a radical of this type if the resulting equation didn't have a trivial solution? – J. C. Nov 21 '16 at 16:00
• The resulting equation in $c$ is a cubic and you are looking for rational roots so there is only a finite number of them to try. Note that not all such radicals can be denested in the general case, see more discussion here and here for example. – dxiv Nov 21 '16 at 16:49
• I was hoping there was a more straight-forward method... Does this work for every "denestable" radical of this type? – J. C. Nov 21 '16 at 19:15
• @JoãoC. I think it works iff the radical is "denestable", and the posts linked in my previous comment would seem to support that, but I don't have a proof handy. – dxiv Nov 21 '16 at 19:34