Answer
$(x^{2}+y^{6})(x^4-x^2y^6+y^{12})(x+y^{3})(x^2-xy^3+y^6)(x-y^{3})(x^2+xy^3+y^6)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
x^{12}-y^{36}
,$ use the factoring of the difference of $2$ squares and the factoring of the sum or difference of $2$ cubes.
$\bf{\text{Solution Details:}}$
Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(x^{6})^2-(y^{18})^2
\\\\=
(x^{6}+y^{18})(x^{6}-y^{18})
\\\\=
(x^{6}+y^{18})[(x^{3})^2-(y^{9})^2]
\\\\=
(x^{6}+y^{18})(x^{3}+y^{9})(x^{3}-y^{9})
.\end{array}
Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
[(x^{2})^3+(y^{6})^3](x^{3}+y^{9})(x^{3}-y^{9})
\\\\=
(x^{2}+y^{6})(x^4-x^2y^6+y^{12})(x^{3}+y^{9})(x^{3}-y^{9})
\\\\=
(x^{2}+y^{6})(x^4-x^2y^6+y^{12})[(x)^{3}+(y^{3})^3](x^{3}-y^{9})
\\\\=
(x^{2}+y^{6})(x^4-x^2y^6+y^{12})(x+y^{3})(x^2-xy^3+y^6)(x^{3}-y^{9})
\\\\=
(x^{2}+y^{6})(x^4-x^2y^6+y^{12})(x+y^{3})(x^2-xy^3+y^6)[(x)^{3}-(y^{3})^3]
\\\\=
(x^{2}+y^{6})(x^4-x^2y^6+y^{12})(x+y^{3})(x^2-xy^3+y^6)(x-y^{3})(x^2+xy^3+y^6)
.\end{array}
