Imaginary numbers

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Min Hoang moderators
16/03/2017 at 10:38
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Let z be one of the imaginary roots of x3 = 27. Evaluate (9 - 3z + z2).(9 + 3z + z2).

Imaginary numbers

  • ...
    FA KAKALOTS 03/02/2018 at 12:38

    Sign A=(9−3z+z2)(9+3z+z2), we have A=(z2+9)2−(3z)2=z4+9z2+92

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    Therefore                (z2−9)A=(z2−9)(z4+9z2+92)=

                                                        =(z2)3−(9)3=(z3)2−(32)3=(27)2−36=(33)2−36=0

    .

    So that  A=0

    So that 

  • ...
    mathlove 16/03/2017 at 11:03

    Sign \(A=\left(9-3z+z^2\right)\left(9+3z+z^2\right)\), we have \(A=\left(z^2+9\right)^2-\left(3z\right)^2=z^4+9z^2+9^2\).

    Therefore                \(\left(z^2-9\right)A=\left(z^2-9\right)\left(z^4+9z^2+9^2\right)=\)

                                                        \(=\left(z^2\right)^3-\left(9\right)^3=\left(z^3\right)^2-\left(3^2\right)^3=\left(27\right)^2-3^6=\left(3^3\right)^2-3^6=0\).

    So that  \(A=0\)

    So that