cubic equation help

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• Oct 15th 2008, 03:26 PM
siegfried
cubic equation help
show that one fo the roots of the equation $x^3-6x^2+9x-1=0$ is $4\sin^2\frac{\pi}{18}$ and find the other two roots
• Oct 15th 2008, 07:04 PM
ThePerfectHacker
Quote:

Originally Posted by siegfried
show that one fo the roots of the equation $x^3-6x^2+9x-1=0$ is $4\sin^2\frac{\pi}{18}$ and find the other two roots

Let $x=y+2$.
This gives, $(y+2)^3 - 6(y+2)^2 + 9(y+2) - 1 = 0$.
Which simplifies to $y^3-3y+1=0$.

Let $C = - \frac{\Delta}{108} = - \frac{3}{4}$

If $A = -\frac{1}{2} + \sqrt{C} = e^{2\pi i/3}$ and $B = -\frac{1}{2} - \sqrt{C} = e^{-2\pi i/3}$.
Then this means $u=\sqrt[3]{A} = e^{2\pi i/9}$ and $v=\sqrt[3]{B} = e^{-2\pi i/9}$ would produce solutions:
$u+v,u\zeta + v\zeta^2,u\zeta^2+v\zeta$ (where $\zeta = e^{2\pi i/3})$

Simplification of these formulas leads to the solutions:
$2\cos \frac{2\pi}{9}, 2\cos \frac{8\pi}{9}, 2\cos \frac{14\pi}{9}$.

And so the three solutions to the original are:
$x_1 = 2 + 2\cos \frac{2\pi}{9} = 4\sin^2 \frac{\pi}{18}$
$x_2 = 2 + 2\cos \frac{8\pi}{9} = 4\sin^2 \frac{4\pi}{18}$
$x_3 = 2 + 2\cos \frac{14\pi}{9}= 4\sin^2 \frac{7\pi}{18}$
• Oct 15th 2008, 08:24 PM
siegfried
thank you theperfecthacker

i believe you are using cardanos method but i think there is an easier method than cardanos
• Oct 15th 2008, 08:30 PM
ThePerfectHacker
Quote:

Originally Posted by siegfried
thank you theperfecthacker

i believe you are using cardanos method but i think there is an easier method than cardanos

It is Cardano's method with a lot of work omitted as well as computations.
• Oct 16th 2008, 03:07 AM
Opalg
Use the formulas $\cos(2\theta) = 1 - 2\sin^2\theta$ and $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$. These show that if $x=4\sin^2\theta$ then $\cos(2\theta) = 1 - \tfrac x2$, and so $\cos(6\theta) = 4\left(1 - \tfrac x2\right)^3 - 3\left(1 - \tfrac x2\right) = 1 - \tfrac92x + 3x^2 - \tfrac12x^3$. But if $\theta = \tfrac\pi{18}$ then $\cos(6\theta)=\tfrac12$. That gives the equation $x^3-6x^2+9x-1=0$. The other two solutions will come from the other angles satisfying $\cos(6\theta)=\tfrac12$, namely $\theta = \tfrac{5\pi}{18}$ and $\theta = \tfrac{7\pi}{18}$. (ThePerfectHacker had $\theta = \tfrac{4\pi}{18}$, but I think $\theta = \tfrac{5\pi}{18}$ must be correct.)
• Oct 16th 2008, 07:25 AM
ThePerfectHacker
Quote:

Originally Posted by Opalg
(ThePerfectHacker had $\theta = \tfrac{4\pi}{18}$, but I think $\theta = \tfrac{5\pi}{18}$ must be correct.)

Yes I made a mistake. But we are going to keep it secret. (Lipssealed)

The mistake was in the simplification. The solutions that I got to the depressed cubic are correct I just made mistakes in the identities.
Here is the corrected version:

1) $2 + 2\cos \frac{2\pi}{9} = 2 - 2\cos \frac{7\pi}{9} = 4\sin^2 \frac{7\pi}{18}$

2) $2 + 2\cos \frac{8\pi}{9} = 2 - 2\cos \frac{\pi}{9} = 4\sin^2 \frac{\pi}{18}$

3) $2 + 2\cos \frac{14\pi}{9} = 2 - 2\cos \frac{5\pi}{9} = 4\sin^2 \frac{5\pi}{18}$