Answer
$(x-1)(x+1)(x^2+1)(x^2+x+1)(x^2-x+1)(x^4-x^2+1)$
Work Step by Step
Using $a^2-b^2=(a+b)(a-b)$ and $a^3-b^3=(a-b)(a^2+ab+b^2)$, then the given expression, $
x^{12}-1
$, is equivalent to
\begin{array}{l}
(x^3)^{4}-1^4
\\\\=
[(x^3)^2-1^2][(x^3)^2+1^2]
\\\\=
(x^6-1)(x^6+1)
\\\\=
(x^3-1)(x^3+1)(x^6+1)
\\\\=
(x-1)(x^2+x+1)(x^3+1)(x^6+1)
\\\\=
(x-1)(x^2+x+1)(x+1)(x^2-x+1)(x^6+1)
\\\\=
(x-1)(x^2+x+1)(x+1)(x^2-x+1)(x^2+1)(x^4-x^2+1)
\\\\=
(x-1)(x+1)(x^2+1)(x^2+x+1)(x^2-x+1)(x^4-x^2+1)
.\end{array}
