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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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@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
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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
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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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