If #x^2 + x +1=0# find the value of #(x+1/x)^3 + (x^2+1/x^2)^3...(x^100+1/x^100)^3#?
3 Answers
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Explanation:
Given:
#x^2+x+1 = 0#
We want to evaluate:
#(x+1/x)^3+(x^2+1/x^2)^3+...+(x^100+1/x^100)^3#
Without loss of generality, let:
#x = omega = -1/2+sqrt(3)/2i#
The only other option is the complex conjugate
Note that:
#omega^3-1 = (omega-1)(omega^2+omega+1) = 0#
So:
#omega^3 = 1#
Let's look at the first few binomials:
#omega+1/omega = omega+bar(omega) = -1#
#omega^2+1/omega^2 = bar(omega)+omega = -1#
#omega^3+1/omega^3 = 1+1 = 2#
These three values repeat cyclically.
So the sum is expressible as:
#(-1-1+8)+(-1-1+8)+...+(-1-1+8)-1 = 33*6-1 = 197#