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" LEAGUE OF LEGENDS "

27/06/2018 at 09:29
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Let x , y , z > 0 .Validity : \(p=\dfrac{x^{2018}+1}{x^{2018}+y^{2018}+z^{2018}+3}\)  know  \(x^3+y^3+z^3=3xyz\)




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    Dao Trong Luan Coordinator 28/06/2018 at 01:00

    \(x^3+y^3+z^3=3xyz\)

    \(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)

    \(\Leftrightarrow\left(x^3+y^3\right)+z^3-3xyz=0\)

    \(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)

    \(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-2\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)

    \(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)

    \(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

    \(x+y+z\ne0\Leftrightarrow x^2+y^2+z^2-xy-yz-zx=0\)

    \(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)

    \(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

    \(\Leftrightarrow x=y=z\)

    \(\Rightarrow p=\dfrac{x^{2018}+1}{x^{2018}+x^{2018}+x^{2018}+3}=\dfrac{x^{2018}+1}{3\left(x^{2018}+1\right)}=\dfrac{1}{3}\)

    " LEAGUE OF LEGENDS " selected this answer.

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