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Math Help - simplification

  1. #1
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    simplification

    z_1^2+z_2^2+z_3^2-z_1z_2-z_2z_3-z_3z_1= (z_1-z_2)(z_2-z_3)+(z_2-z_3)(z_3-z_1)+(z_3-z_1)(z_1-z_2)plz discuss how the cyclic symmetry works in the factorization?
    Last edited by Navesh; June 19th 2006 at 06:40 AM.
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  2. #2
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    Re: simplification

    Solution:
    Given f(z1,z2,z3)=z1+z2+z3-z1z2-z2z3-z3z1=(z1-z2)(z2-z3)+(z2-z3)(z3-z1)+(z3-z1)
    (z1-z2)
    consider f(z2,z3,z1)=z2+z3+z1-z2z3-z3z1-z1z2=(z2-z3)(z3-z1)+(z3-z1)(z1-z2)+(z1-z2)(z2-z3)
    f(z3,z1,z2)=z3+z1+z2-z3z1-z1z2-z2z3-=(z3-z1)(z1-z2)+(z1-z2)(z2-z3)+(z2-z3)(z3-z1)
    here f(z1,z2,z3)=f(z2,z3,z1)=f(z3,z1,z2)
    there for the factorization is cyclic symmetric.


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