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Math Help - cubic equation help

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by siegfried View Post
    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}
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  3. #3
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    thank you theperfecthacker

    i believe you are using cardanos method but i think there is an easier method than cardanos
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    Quote Originally Posted by siegfried View Post
    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.
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  5. #5
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    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.)
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  6. #6
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    Quote Originally Posted by Opalg View Post
    (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.

    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}
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