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ThePerfectHacker · #2 October 15th, 2008, 09:04 PM

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:
$x_1 = 2 + 2\cos \frac{2\pi}{9} = 4\sin^2 \frac{\pi}{18}$
$x_2 = 2 + 2\cos \frac{8\pi}{9} = 4\sin^2 \frac{4\pi}{18}$
$x_3 = 2 + 2\cos \frac{14\pi}{9}= 4\sin^2 \frac{7\pi}{18}$

The following users thank ThePerfectHacker for this useful post:
siegfried
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