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Re: #p#

by parallel_chase » Fri Aug 29, 2008 3:59 pm
ddm wrote:attachment
#p#=ap^3 + bp -1

#-7#= -343a - 7b -1

-343a - 7b -1 = 3

343a + 7b +1 = -3

343a + 7b = -4

#7# = 343a + 7b -1

#7# = -4 -1 = -5

Hope this helps.
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