Factorization of Polynomial with complex coefficients of the form $n^6-a$ into the form of $(n^2-a_1)(n^2-a_2)(n^2-a_3)$

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In a textbook that I'm reading (in Complex Analysis) the following equation is presented



$n^6-8=(n^2-2)(n^2-2w)(n^2-2w^2)$



where $w$ is some root of order 3 of 1 (meaning $w^3=1$) (eg. $w=-frac12 + fracsqrt32i$).



I am wondering what formula / method can be used to factorize a polynomial with complex coefficients of the form



$n^6-a$



into the form of



$(n^2-a_1)(n^2-a_2)(n^2-a_3)$



Thanks!




polynomials complex-numbers




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    up vote
    2
    down vote

    favorite












    In a textbook that I'm reading (in Complex Analysis) the following equation is presented



    $n^6-8=(n^2-2)(n^2-2w)(n^2-2w^2)$



    where $w$ is some root of order 3 of 1 (meaning $w^3=1$) (eg. $w=-frac12 + fracsqrt32i$).



    I am wondering what formula / method can be used to factorize a polynomial with complex coefficients of the form



    $n^6-a$



    into the form of



    $(n^2-a_1)(n^2-a_2)(n^2-a_3)$



    Thanks!




    polynomials complex-numbers




    share|cite|improve this question





















      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      In a textbook that I'm reading (in Complex Analysis) the following equation is presented



      $n^6-8=(n^2-2)(n^2-2w)(n^2-2w^2)$



      where $w$ is some root of order 3 of 1 (meaning $w^3=1$) (eg. $w=-frac12 + fracsqrt32i$).



      I am wondering what formula / method can be used to factorize a polynomial with complex coefficients of the form



      $n^6-a$



      into the form of



      $(n^2-a_1)(n^2-a_2)(n^2-a_3)$



      Thanks!




      polynomials complex-numbers




      share|cite|improve this question











      In a textbook that I'm reading (in Complex Analysis) the following equation is presented



      $n^6-8=(n^2-2)(n^2-2w)(n^2-2w^2)$



      where $w$ is some root of order 3 of 1 (meaning $w^3=1$) (eg. $w=-frac12 + fracsqrt32i$).



      I am wondering what formula / method can be used to factorize a polynomial with complex coefficients of the form



      $n^6-a$



      into the form of



      $(n^2-a_1)(n^2-a_2)(n^2-a_3)$



      Thanks!





      polynomials complex-numbers





      share|cite|improve this question










      share|cite|improve this question




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      asked Jul 30 at 6:24









      jackjack

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          Hint:   $z^3-1 = (z-1)(z-w)(z-w^2),$, then let $,z = n^2/2,$ or, in general, $,z=n^2/sqrt[3]a,$.






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            Hint:   $z^3-1 = (z-1)(z-w)(z-w^2),$, then let $,z = n^2/2,$ or, in general, $,z=n^2/sqrt[3]a,$.






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              up vote
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              Hint:   $z^3-1 = (z-1)(z-w)(z-w^2),$, then let $,z = n^2/2,$ or, in general, $,z=n^2/sqrt[3]a,$.






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                up vote
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                up vote
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                accepted






                Hint:   $z^3-1 = (z-1)(z-w)(z-w^2),$, then let $,z = n^2/2,$ or, in general, $,z=n^2/sqrt[3]a,$.






                share|cite|improve this answer













                Hint:   $z^3-1 = (z-1)(z-w)(z-w^2),$, then let $,z = n^2/2,$ or, in general, $,z=n^2/sqrt[3]a,$.







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                answered Jul 30 at 6:28









                dxiv

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