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Math Help - Simplifying Ugly Radical Expressions from Cardano's Method

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    Simplifying Ugly Radical Expressions from Cardano's Method

    I asked a similar question to this some time ago, but as we have gained several new specialists since then, I thought I'd try again.

    The problem comes from solving the equation
    x^3 - 7x + 6 = 0
    by Cardano's method.

    I get an expression for x of the form:
    x = 2\sqrt[3]{\frac{7 \sqrt{21}}{9}}cos \left ( \frac{1}{3} atn \left ( \frac{10\sqrt{3}}{27} \right ) \right )

    As it happens,
    2\sqrt[3]{\frac{7 \sqrt{21}}{9}}cos \left ( \frac{1}{3} atn \left ( \frac{10\sqrt{3}}{27} \right ) \right ) = 3

    but how can I prove that? Is there any way to simplify this monster without knowing the answer ahead of time?

    -Dan
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    I am not sure what you are asking.
    But sometimes I get expressions which look like,

    \sqrt[3]{1+\sqrt{2}}+\sqrt[3]{1-\sqrt{2}}.

    And in general there is no way to simplify it, you just leave it like that. Part of the difficulty in the word "simplify" is that it is not well-defined, and hence we cannot come up with a theory which tells us when we can "simplify" and when we cannot. Perhaps you can create a definition.

    And, I also do not understand how you got trigonometric functions in your answer. I never remember learning it that.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I am not sure what you are asking.
    But sometimes I get expressions which look like,

    \sqrt[3]{1+\sqrt{2}}+\sqrt[3]{1-\sqrt{2}}.

    And in general there is no way to simplify it, you just leave it like that. Part of the difficulty in the word "simplify" is that it is not well-defined, and hence we cannot come up with a theory which tells us when we can "simplify" and when we cannot. Perhaps you can create a definition.

    And, I also do not understand how you got trigonometric functions in your answer. I never remember learning it that.
    The answer is that I would like to see if there is a way I can know that
    2\sqrt[3]{\frac{7 \sqrt{21}}{9}}cos \left ( \frac{1}{3} atn \left ( \frac{10\sqrt{3}}{27} \right ) \right ) = 3
    without resorting to a decimal approximation.

    I did make a mistake in my derivation. (That's what I get for not writing it out in full.)

    As to how I got that form:
    x^3 - 7x + 6 = 0

    Let x = p - q and q = -\frac{7}{3p}

    Then

    p^3 + \frac{343}{27p^3} + 6 = 0

    27p^6 + 162p^3 + 343 = 0

    p^3 = -3 + i \frac{10\sqrt{3}}{9} is one solution.

    Now put p^3 into complex polar form. (This will "simplify" the eventual answer for x.)

    p^3 = \frac{7\sqrt{21}}{9} \cdot e^{i \cdot atn \left ( \frac{9\sqrt{3}}{10} \right ) + i \frac{ \pi}{2}}

    So
    p = \sqrt[3]{\frac{7\sqrt{21}}{9}} e^{\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) + i \frac{\pi}{6}}


    Thus
    q = -\frac{7}{3p} = -\frac{7}{3} \sqrt[3]{\frac{9}{7\sqrt{21}}}<br />
e^{-\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) - i \frac{\pi}{6}}

    q = -\sqrt[3]{\frac{7\sqrt{21}}{9}} e^{-\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) - i \frac{\pi}{6}}

    Thus
    x = p - q = \sqrt[3]{\frac{7\sqrt{21}}{9}} e^{\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) + i \frac{\pi}{6}} + \sqrt[3]{\frac{7\sqrt{21}}{9}} e^{-\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) - i \frac{\pi}{6}}

    x = 2 \sqrt[3]{\frac{7\sqrt{21}}{9}} \left ( \frac{ e^{\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) + i \frac{\pi}{6}} + <br />
e^{-\frac{i}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) - i \frac{\pi}{6}}}{2} \right )<-- This is the reason for the complex polar form

    x = 2 \sqrt[3]{\frac{7\sqrt{21}}{9}} cos \left ( <br />
\frac{1}{3} atn \left ( \frac{9\sqrt{3}}{10} \right ) + \frac{\pi}{6} \right )

    x = 2

    -Dan
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