Factorization of $(1+x+x^2+x^3)^2 - x^3$

Question

Factorize : $(1+x+x^2+x^3)^2 - x^3$

I've tried to expand it but the equation will be even more complicated, anyone can give me some hints to solve it without expanding it (or it is necessary to expand it)?

Answer 1

A strange but efficient way that uses the geometric series formula 3 times:

$$(1+x+x^2+x^3)=\frac{1-x^4}{1-x}$$ Your polynomial is then: $$p(x)=(1+x+x^2+x^3)^2-x^3=\frac{(1-x^4)^2-x^3(1-x)^2}{(1-x)^2}$$ $$=\frac{1-2x^4+x^8-x^3+2x^4-x^5}{(1-x)^2}=\frac{1-x^3-x^5+x^8}{(1-x)^2}=$$ $$=\frac{(1-x^3)(1-x^5)}{(1-x)(1-x)}=(1+x+x^2)(1+x+x^2+x^3+x^4)$$

Answer 2

$(1+x+x^2+x^3)^2-x^3 = (1+x+x^2)^2 + 2x^3(1+x+x^2) + x^3(x-1)(1+x+x^2) = (1+x+x^2)(1+x+x^2+x^3+x^4)$