For what values of m(m ∈ Z, m >= 0) is polynomial $\displaystyle (x+1)^m-x^m-1$ dividable with $\displaystyle (x^2+x+1)^3$
Last edited by Garas; Dec 11th 2010 at 01:59 AM.
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Wite $\displaystyle (x^2+ x+ 1)^3$ as $\displaystyle ((x+ 1)+ x^2)^3= (x+1)^3+ 3x^2(x+1)^2+ 3x^4(x+1)+ x^6$ The first thing that occurs to me is that m should be a multiple of 3.
No, it's not so simple ((x+1)^9-x^9-1)/((x^2+x+1) 1; - Wolfram|Alpha or ((x+1)^6-x^6-1)/((x^2+x+1) 1; - Wolfram|Alpha
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