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    How do i find the roots of this eqn?

    Q: Find the eqn whose roots are the fourth powers of the roots of the eqn x^3 + x + 1 = 0

    I understand the question, just cant see how to get the roots
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    You don't need to find the roots -- and in fact two of the roots are complex

    Call the roots of that equation \alpha, \beta and \gamma.

    So \alpha + \beta + \gamma = 0
    \alpha \beta + \alpha \gamma + \beta \gamma = 1
    \alpha \beta \gamma = -1

    You need to use combinations of these to find \alpha^4 + \beta^4 + \gamma^4 and \alpha^4 \beta^4 + \alpha^4 \gamma^4 + \beta^4 \gamma^4 and \alpha^4 \beta^4 \gamma^4 (the last one is easy enough).

    Then you can form the equation.
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    lots of ways,
    substitute  x = w + \frac{m}{w}
    and choose m so that you get only terms in w^3, w^0 and w^(-3)?

    edit: take Glutamic acid's answer, I was purely trying to find the roots;
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    I think a substitution method might be quicker than Glutamic Acid's method.

    x^4=u

    x=u^{1/4}

    Rearrange and do some squaring and you get the answer pretty quickly.
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    I have never seen the substitution method explained to my satisfaction in any book (probably because I'm a bit simple )..so I'm going to try here with this as an example.

    Spoiler:
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    If \alpha, \beta  \text{ and } \gamma are the roots of

    x^3+x+1=0 then \alpha^3+\alpha+1\equiv 0

    and if u=\alpha^4 then u^{1/4}=\alpha

    and so

    u^{3/4}+u^{1/4}+1\equiv 0

    Rearranging ...

    u^{3/4}+u^{1/4}\equiv -1

    Squaring

    u^{3/2}+2u+u^{1/2}\equiv 1

    Rearranging

    u^{3/2}+u^{1/2}\equiv 1-2u

    Squaring

    u^3+2u^{2}+u \equiv 1-4u+4u^2

    and rearranging again

    u^3-2u^{2}+5u-1 \equiv 0

    and so a cubic with the required roots is

    x^3-2x^{2}+5x-1 = 0



    I'm sure this isn't perfect and someone out there can do better. If so please do.
 
 
 
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