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imqwerty

  • one year ago

another ques..

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  1. imqwerty
    • one year ago
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    l,m,n are positive real numbers such that\[l^3+m^3=n^3\]prove that\[l^2+m^2-n^2>6(n-l)(n-m)\]

  2. MrNood
    • one year ago
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    I am not sure about this - but I was under the impression that there were NO solutions for that equation (isn't this Fermat's last theorem problem?)

  3. ganeshie8
    • one year ago
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    Fermat's last theorem was about "integer" solutions

  4. mathmath333
    • one year ago
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    \(\color{white}{bookmark}\)

  5. AbdullahM
    • one year ago
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    Didn't @kainui solve this yesterday?

  6. imqwerty
    • one year ago
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    thas ws not the proper solution :)

  7. MrNood
    • one year ago
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    @ganeshie8 ah yes - sorry my mistake

  8. imqwerty
    • one year ago
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    ok :) heres the solution-

  9. thadyoung
    • one year ago
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  10. imqwerty
    • one year ago
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    ( ͡° ͜ʖ ͡°)=found

  11. ParthKohli
    • one year ago
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    That's a crazy solution. You can't really ever normally think of this.

  12. imqwerty
    • one year ago
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    solution obtained by normal thinking - u can write that inequality like -\[l^2+m^2-n^2-6(n-l)(n-m)\]\[7n^2-6(l+m)n-(l^2+m^2-6lm)<0\] x=7c^2 y=-6(l+m)n z=-(l^2 +m^2 -6lm) so u have to prove x+y+z<0 u can also observe that x ,y, z are not all equal x>0 , y<0 use the identity- \[x^3+y^3+z^3-3xyz=\frac{ 1 }{ 2 }(x+y+z)[(x-y)^2 +(y-z)^2+(z-x)^2]\] but its enough to prove that x^2 + y^3+z^3 -3xyz<0 substitute x, y,z :/ i didn't tell this cause i hate the tedious calculations it involves after some simplification this will reduce to-\[-l^2m^2(129l^2+129m^2-254lm)<0\] notice that this thing does is not having n cause i substituted n^3=m^3+l^3 while simplifying and clearly \[129l^2 +129m^2 -254lm=129(l-m)^2+4lm>0 \] hence the results follow ...

  13. ParthKohli
    • one year ago
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    haha normal thinking

  14. imqwerty
    • one year ago
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    :D

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