Question

another ques..

Answer 1

l,m,n are positive real numbers such that\[l^3+m^3=n^3\]prove that\[l^2+m^2-n^2>6(n-l)(n-m)\]

Answer 2

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?)

Answer 3

Fermat's last theorem was about "integer" solutions

Answer 4

\(\color{white}{bookmark}\)

Answer 5

Didn't @kainui solve this yesterday?

Answer 6

thas ws not the proper solution :)

Answer 7

@ganeshie8
ah yes - sorry my mistake

Answer 8

ok :) heres the solution-

Answer 9

( ͡° ͜ʖ ͡°)=found

Answer 10

That's a crazy solution. You can't really ever normally think of this.

Answer 11

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 ...

Answer 12

haha normal thinking

Answer 13

:D