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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Jan 2008 13:27:37 IST
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(2) x3-x2-x-1 has a root a,b,c than prove that a1992-b1992/a-b + b1992-c1992/b-c + c1992-a1992/c-a is an integer. (3) solve [j=1 ] [ 3] xj /a i-aj=1 (where i=1,2,3)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Jan 2008 16:17:55 IST
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2 find a b c
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Jan 2008 19:15:16 IST
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a^2-b^2 (a-b)(a+b)is divisible by a-b a^3-b^3 (a-b)(a^2+ab+b^2)is divisible by a-b similarly a^4-b^4 is also divisible by a-b it can be thus shown that a^n-b^n is always divisble by a-b for natural number n thus a^1992-b^1992/a-b is an integer similarly the other two so int+int+int which is also an integer
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Jan 2008 21:33:52 IST
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Shoosy is wong !!!! a, b, c arenot integers !! ( Two of them are infact complex )
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Jan 2008 21:41:15 IST
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i think a, b and c are all real.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Jan 2008 21:59:04 IST
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I didnt read the first line my friends so its my miss anyway as far as i know mr feynmann is right f(1)<0 f(2)>0 and by checking up we find only one root (real) so i think there is only 1 real root pls correct if wrong
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Jan 2008 12:13:56 IST
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This doesn't look good at all to me. The cubic equation has only one real root, namely . It has a pair of complex roots . Notice that is large (roughly ), while and are small (roughly ).
In the expression , the two terms in will tend to swamp the others. But is real, and has a substantial non-real component, since . But i dont see how the expression could possibly be a real integer.
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
JEE and OLYMPIA INFINATUM
http://iit-redefined.theforum.name/index.php
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