Respuesta :

LammettHash
Because [tex]w[/tex] is a cube root of unity, you have [tex]w^3=1[/tex]. So

[tex]w^{49}=w(w^3)^{16}=w[/tex]
[tex]w^{101}=w^2(w^3)^{33}=w^2[/tex]
[tex]w^{150}=(w^3)^{50}=1[/tex]

and so

[tex]w^{49}+w^{101}+w^{150}=1+w+w^2=\dfrac{1-w^3}{1-w}[/tex]

provided that [tex]w\neq1[/tex]. Any other cube root of unity will make the numerator vanish, so the equality holds.

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