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Question Number 171972 by mathlove last updated on 22/Jun/22

Answered by mr W last updated on 22/Jun/22

$$x\ne 1\Rightarrow x-1\ne 0$$$$(x-1)(x^4+x^3+x^2+x+1)=0$$$$x^5-1=0$$$$\Rightarrow x^5=1$$$$x^{2010}+x^{2020}={\left(x^5\right)}^{402}+{\left(x^5\right)}^{404}=1^{402}+1^{404}=2$$

Commented by mathlove last updated on 22/Jun/22

$$thanks$$

Commented by infinityaction last updated on 22/Jun/22

$$useG.Psum$$$$\frac{x^5-1}{x-1}=0$$$$x^5=1$$$${\left(x^5\right)}^{402}+{\left(x^5\right)}^{404}=1+1=2$$

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