(x-1)(x+1)=x^2-1,(x-1)(x^2+x+1)=x^3-1.已知x^4+x^3+x^2+x+1=0,求x^2010

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(x-1)(x+1)=x^2-1,(x-1)(x^2+x+1)=x^3-1.已知x^4+x^3+x^2+x+1=0,求x^2010(x-1)(x+1)=x^2-1,(x-1)(x^2+x+1)=x^3

(x-1)(x+1)=x^2-1,(x-1)(x^2+x+1)=x^3-1.已知x^4+x^3+x^2+x+1=0,求x^2010
(x-1)(x+1)=x^2-1,(x-1)(x^2+x+1)=x^3-1.已知x^4+x^3+x^2+x+1=0,求x^2010

(x-1)(x+1)=x^2-1,(x-1)(x^2+x+1)=x^3-1.已知x^4+x^3+x^2+x+1=0,求x^2010
因为(x-1)(x^4+x^3+x^2+x+1) = x^5 - 1 = 0
所以x^5 = 1
故x^2010 = x^(5×402) = (x^5)^402 = 1^402 = 1
希望有用.