The roots of the equation x3 - x + 1 = 0 are a, b, c. Find a8 + b8 + c8.
Answer
10
Solution
We have a+b+c = 0, ab+bc+ca = -1, abc = -1.
a2+b2+c2 = (a+b+c)2 - 2(ab+bc+ca) = 2. Similarly, a2b2+b2c2+c2a2 = (ab+bc+ca)2 - 2abc(a+b+c) = 1. Hence a4+b4+c4 = (a2+b2+c2)2 - 2(a2b2+b2c2+c2a2) = 2. Similarly, a4b4+b4c4+c4a4 = (a2b2+b2c2+c2a2)2 - 2a2b2c2(a2+b2+c2) = 1 - 4 = -3. Finally a8+b8+c8 = (a4+b4+c4)2 - 2(a4b4+b4c4+c4a4) = 4 + 6 = 10.
Thanks to Suat Namli
© John Scholes
jscholes@kalva.demon.co.uk
6 March 2004
Last corrected/updated 6 Mar 04