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Math Help - Existence proofs

  1. #1
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    Existence proofs

    I need to show that there exists no non zero real numbers a,b such that the SqRt of a^2 plus b^2 is equal to the CubeRt of a^3 plus b^3... I have been pluggin in everything cant seem to find the right combination !
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    Hello,
    Quote Originally Posted by nikie1o2 View Post
    I need to show that there exists no non zero real numbers a,b such that the SqRt of a^2 plus b^2 is equal to the CubeRt of a^3 plus b^3... I have been pluggin in everything cant seem to find the right combination !
    (\sqrt{a^2+b^2})^6=(\sqrt[3]{a^3+b^3})^6

    (a^2+b^2)^3=(a^3+b^3)^2

    a^6+b^6+3a^2b^4+3a^4b^2=a^6+b^6+2a^3b^3

    3a^2b^4+3a^4b^2-2a^3b^3=0

    a^2b^2(3b^2-2ab+3a^2)=0

    since a and b are not equal to 0, this is equivalent to :
    3b^2-2ab+3a^2=0

    (\sqrt{3} b)^2-2\cdot \frac{a}{\sqrt{3}} \cdot (\sqrt{3} b)+3a^2=0

    \left(\sqrt{3} b-\frac{a}{\sqrt{3}}\right)^2-\frac{a^2}{3}+3a^2=0

    \left(\sqrt{3} b-\frac{a}{\sqrt{3}}\right)^2+\frac{10}{3} \cdot a^2=0

    this is true if and only if \sqrt{3} b-\frac{a}{\sqrt{3}}=0 and a=0, because both terms in the sum are positive.

    And this is not possible since a \neq 0
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