Question 390481
A neat shortcut:


If you know that the sum of a geometric sequence


{{{x + x^2 + x^3}}} + ... + {{{x^n}}} is {{{(x^(n+1) - 1)/(x-1)}}} we can obtain


{{{x + x^2 + x^3 + x^4 = (x^5 - 1)/(x-1)}}}


Multiplying both sides by x-1, we get the factorization of {{{x^5 - 1}}}


{{{x^5 - 1 = (x + x^2 + x^3 + x^4)(x-1)}}}