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(ax^3+bx^2+cx+d) + (ay^3+by^2+cy+d) = (az^3+bz^2+cz+d)
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January 10th, 2017, 05:43 PM
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Denis
Math Team
Joined: Oct 2011
From: Ottawa Ontario, Canada
Posts: 11,091
Thanks: 723
Quote:
Originally Posted by
Naveenchandra Kumar
It is (5,7,
, .... (117, 228, 23
.
Last edited by Naveenchandra Kumar; January 10th, 2017 at 08:34 PM. Reason: Somehow smiley is sitting over 8! always.
If last digit in brackets is 8, leave a space: (238 )
Thanks from
Naveenchandra Kumar
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multiple of a number

Factorization . Is this possible ?
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