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[Post New]6 Mar 2008 20:08:32 IST
    Subject: Re:Roots of Polynomial
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hsbhatt (4910)

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Olaaa!! Perrrfect answer. 926  [1066 rates]
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Just another method:
Since a3+3a+3 = 0 etc.,
 
We have a5+3a3+3a2=0 etc.(multiplying by a2)
 
Hence a5+3a3+3a2 = 0
 
Also a3 = 3abc = -9 (Since a =0)
 
Now a2 = (a+b+c)2 - 2ab = -6
 
Hence a5 = 3*9+3*6 = 45
Time wounds all heels
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