Respuesta :

LammettHash

Recall that

[tex]1-x^n=(1-x)(1+x+x^2+\cdots+x^{n-1})[/tex]

So we have

[tex]x^2+x+1=0\implies\dfrac{1-x^3}{1-x}=0\implies x^3=1[/tex]

Then for any [tex]n[/tex], we have

[tex]x^{3n}=(x^3)^n=1^n=1[/tex]

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