maths surd

(2+root(3))^6+[(2-root(3)]
without using calcultor

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asked by philip
  1. let
    a=2+sqrt(3) b=2-sqrt(3)
    but
    a=((2+sqrt(3)^2)
    b=[(2-sqrt(3)^2]
    a^2=(2+sqrt(3))^2
    =(2+sqrt(3)(2+sqrt(3)=7+4sqrt(3)
    b^2=(2-sqrt(2)]^2
    =(2-sqrt(3))(2-sqrt(3)
    =7-4sqrt(3)
    but
    (a^3+b^3)=(a+b)(a^2-ab+b^2)
    a+b=7+4sqrt(3)+7-4sqrt(3)
    a+b=14
    ab=(7-4sqr(3)(7+4sqrt(3)
    ab=1
    a^2=(7+4sqrt(3)(7+4sqrt(3)
    =97+56sqrt(3)
    b^2=(7-4sqrt(3)^2
    b^2=(7-4sqrt(3)(7-4sqrt(3)=97-56sqrt(3)
    a^3+b^3=(a+b)(a^2-ab+b^2)
    =14(97+56sqrt(3)-1+96-56sqrt(3)
    =14(97-1+97)
    =14(193)=2702 yeah

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    posted by collins

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