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Prove that `x^3 + x^2 + x` is factor of `(x+1)^n - x^n -1` where n is odd integer greater than 3, but not a multiple of 3. |
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Answer» `x^3+x^2+x=(x+1)-x^2-1` `x[1+x+x^2]` `1+w+w^2=0` `x(x-w)(x-w^2)` When x=0 `(0+1)^n-(0)^n-1=0` when x=w `(1+w)^n-w^n-1` `=(-w^2)^n-w^n-1` `=(-w)^n-(w)^n-(w^3)^n` `1+w+w^2=0` `1+w^2+w^(2n)=0` Only when n is odd n is not multiple of 3 `(-w)^n-(w^2)^n-1` n is odd. |
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