# Solving a cubic equation

I have a math problem in front of me, and after struggling three days with it, I cannot find the solution. I need to find the roots of this polynomial equation: $x^3+x^2+1=0$.

Anyone have a solution to this?

Any help is appreciated..

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– Sean Eberhard Sep 10 '12 at 16:16
@SeanEberhard I did that, however I am interested in the steps to the solution. – speci Sep 10 '12 at 16:19
I would use Tartaglia-Cardano method, what is a lot of work. – Integral Sep 10 '12 at 16:19
You could proceed as follows: set $y=\frac 1 x$ so that $y^3+y+1=0$. Then set $y=u+v$ with $uv=-\frac 1 3$ which enables you to obtain expressions for $u^3+v^3$ and $u^3v^3$. Then you can solve a quadratic to get $u^3$ and $v^3$, which gives you $u$ and $v$ and hence $y$ and $x$. – Mark Bennet Sep 10 '12 at 16:34

## 1 Answer

Solved the equation using Tartaglia's method $$x^3 + x^2 + 1$$ Make then substitution $x = t + h$ $$t^3 + 3t^2h+3th^2 + h^3 + t^2 +2th+h^2 + 1 = 0$$ $$t^3 + t^2(3h+1)+t(3h^2+2h) + (h^3 + h^2 + 1) = 0$$

In order to eliminate the second degree term we add a condition $h = -\frac{1}{3}$

$$t^3 -\frac{1}{3}t + \frac{29}{27}$$

We make $t = u+v$

$$u^3 + 3u^2 v + 3uv^2 + v^3 -\frac{1}{3}u - \frac{1}{3}v + \frac{29}{27}=0$$ $$u^3 + v^3 + \frac{29}{27} + 3u^2 v -\frac{1}{3}u + 3uv^2 - \frac{1}{3}v =0$$ $$u^3 + v^3 + \frac{29}{27} + u\left(3u v -\frac{1}{3}\right) + v\left(3uv - \frac{1}{3}\right) =0$$ $$u^3 + v^3 + \frac{29}{27} + (u+v)\left(3u v -\frac{1}{3}\right) =0$$

We make a system of equation $$u^3 + v^3 + \frac{29}{27} = 0; (u+v)\left(3u v -\frac{1}{3}\right) =0$$

For (2) we have: $u \ne v$ because does not satisfy (1), then $$u = \frac{1}{9v} \rightarrow u^3 = \frac{1}{729v^3}$$

For (1) we have: $$u^3 + v^3 + \frac{29}{27} = 0$$ $$\frac{1}{729v^3} + v^3 + \frac{29}{27} = 0$$ $$\left(v^3\right)^2 + \frac{29v^3}{27} + \frac{1}{729} = 0$$

Using general quadratic formula:

$$v^3 = \frac{-\frac{29}{27} \pm \sqrt{\left(\frac{29}{27} \right)^2 - 4\frac{1}{729} }}{2}$$ $$v^3 = -\frac{29}{54} \pm \frac{1}{2}\sqrt\frac{31}{27}$$

We hope $u^3$ to be the conjugate of $v^3$ because they are interchangeable. (Try solving $v^3$ to prove it).

So, $$u+v= \sqrt[3]{-\frac{29}{54} + \frac{1}{2}\sqrt\frac{31}{27}} + \sqrt[3]{-\frac{29}{54} - \frac{1}{2}\sqrt\frac{31}{27}}$$

So, we can now have the value of $x$: $$x = t + h = u + v + h = \sqrt[3]{-\frac{29}{54} + \frac{1}{2}\sqrt\frac{31}{27}} + \sqrt[3]{-\frac{29}{54} - \frac{1}{2}\sqrt\frac{31}{27}} - \frac{1}{3}$$

$$x \approx -1.46557$$

Good luck finding the exact complex solutions ._.

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