4

I'm trying to calculate the exact values of the roots of $x^3 - x^2 +1 =0$ (When I say exact value, here I mean find the roots in the form of $a + bi$ for some $a,b \in \mathbb{R}$

My first thought is to get it to a form where I can use Cardano's method and so, since 0 isn't a root, I took $x = 1/t$ and then got

$t^3 - t +1 = 0$

So by Cardano's formula, I got that

t = $\sqrt[3]{ \frac{-1}{2} + \sqrt[2]{\frac{1}{4}-\frac{1}{27}}} \sqrt[3]{ \frac{-1}{2} - \sqrt[2]{\frac{1}{4}-\frac{1}{27}}}$

= $\sqrt[3]{ \frac{-1}{2} + \sqrt[2]{\frac{23}{108}}} + \sqrt[3]{ \frac{-1}{2} - \sqrt[2]{\frac{23}{108}}}$

In $\mathbb{C}$, both $\sqrt[3]{ \frac{-1}{2} + \sqrt[2]{\frac{23}{108}}}$ and $\sqrt[3]{ \frac{-1}{2} - \sqrt[2]{\frac{23}{108}}}$ have three cube roots, and so taking $u$ and $v$ to be some particular cube roots, I know that I need to compute $\omega ^k u + \omega^l v$ for $k,l \in \{0,1,2\}$ and $\omega$ being a cube root of unity.

But I'm not sure about 1) how to compute any particular values of $u$ and $v$ in this case, as well as 2) if there is a cleaner way to approach this problem and solving monic cubic polynomials of this form in general.

Any advice would be appreciated!

NoChance
  • 6,695
User20354
  • 1,070
  • https://math.stackexchange.com/questions/2157643/how-can-i-solve-the-equation-x3-x-1-0/2157645#2157645 – lab bhattacharjee Sep 13 '19 at 01:26
  • See also: https://math.stackexchange.com/questions/2203364/solve-the-following-equation-x3-3x-sqrtx2 – lab bhattacharjee Sep 13 '19 at 01:28
  • I don't think the term "exact" here as you define it is correct. You could simply say you want the values of a, b in radical form, because "exact" has an exact meaning! – NoChance Sep 13 '19 at 04:50
  • Right. Apologies for the ambiguity. I meant that I wanted to find solutions in that form (which I probably incorrectly called exact) – User20354 Sep 16 '19 at 18:09

1 Answers1

2

Without change of variables, using Cardano method, the raw results are $$x_1=\frac{1}{3} \left(1-\sqrt[3]{\frac{2}{25-3 \sqrt{69}}}-\sqrt[3]{\frac{1}{2} \left(25-3 \sqrt{69}\right)}\right)$$ $$x_2=\frac{1}{3}+\frac{1}{6} \left(1+i \sqrt{3}\right) \sqrt[3]{\frac{1}{2} \left(25-3 \sqrt{69}\right)}+\frac{1-i \sqrt{3}}{3\ 2^{2/3} \sqrt[3]{25-3 \sqrt{69}}}$$ $$x_3=\frac{1}{3}+\frac{1}{6} \left(1-i \sqrt{3}\right) \sqrt[3]{\frac{1}{2} \left(25-3 \sqrt{69}\right)}+\frac{1+i \sqrt{3}}{3\ 2^{2/3} \sqrt[3]{25-3 \sqrt{69}}}$$

Now, being patient $$x_{2,3}=a\pm b i$$ where $$a=\frac{1}{3}+\frac{1}{6} \sqrt[3]{\frac{1}{2} \left(25-3 \sqrt{69}\right)}+\frac{1}{3\ 2^{2/3} \sqrt[3]{25-3 \sqrt{69}}}$$ $$b=\frac{\sqrt[3]{\frac{1}{2} \left(25-3 \sqrt{69}\right)}}{2 \sqrt{3}}-\frac{1}{2^{2/3} \sqrt{3} \sqrt[3]{25-3 \sqrt{69}}}$$

Claude Leibovici
  • 289,558