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imqwerty
 one year ago
another ques..
imqwerty
 one year ago
another ques..

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imqwerty
 one year ago
Best ResponseYou've already chosen the best response.3l,m,n are positive real numbers such that\[l^3+m^3=n^3\]prove that\[l^2+m^2n^2>6(nl)(nm)\]

MrNood
 one year ago
Best ResponseYou've already chosen the best response.0I 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?)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Fermat's last theorem was about "integer" solutions

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{white}{bookmark}\)

AbdullahM
 one year ago
Best ResponseYou've already chosen the best response.0Didn't @kainui solve this yesterday?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.3thas ws not the proper solution :)

MrNood
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 ah yes  sorry my mistake

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.3ok :) heres the solution

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0That's a crazy solution. You can't really ever normally think of this.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.3solution obtained by normal thinking  u can write that inequality like \[l^2+m^2n^26(nl)(nm)\]\[7n^26(l+m)n(l^2+m^26lm)<0\] 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 \[x^3+y^3+z^33xyz=\frac{ 1 }{ 2 }(x+y+z)[(xy)^2 +(yz)^2+(zx)^2]\] 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\[l^2m^2(129l^2+129m^2254lm)<0\] notice that this thing does is not having n cause i substituted n^3=m^3+l^3 while simplifying and clearly \[129l^2 +129m^2 254lm=129(lm)^2+4lm>0 \] hence the results follow ...

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0haha normal thinking
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