Answer
$(a^{2}+b^{6})(a^4-a^2b^6+b^{12})(a+b^{3})(a^2-ab^3+b^6)(a-b^{3})(a^2+ab^3+b^6)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
a^{12}-b^{36}
,$ use the factoring of the difference of $2$ squares and the factoring of the sum or difference o $2$ 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}
(a^{6})^2-(b^{18})^2
\\\\=
(a^{6}+b^{18})(a^{6}-b^{18})
\\\\=
(a^{6}+b^{18})[(a^{3})^2-(b^{9})^2]
\\\\=
(a^{6}+b^{18})(a^{3}+b^{9})(a^{3}-b^{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}
[(a^{2})^3+(b^{6})^3](a^{3}+b^{9})(a^{3}-b^{9})
\\\\=
(a^{2}+b^{6})(a^4-a^2b^6+b^{12})(a^{3}+b^{9})(a^{3}-b^{9})
\\\\=
(a^{2}+b^{6})(a^4-a^2b^6+b^{12})[(a)^{3}+(b^{3})^3](a^{3}-b^{9})
\\\\=
(a^{2}+b^{6})(a^4-a^2b^6+b^{12})(a+b^{3})(a^2-ab^3+b^6)(a^{3}-b^{9})
\\\\=
(a^{2}+b^{6})(a^4-a^2b^6+b^{12})(a+b^{3})(a^2-ab^3+b^6)[(a)^{3}-(b^{3})^3]
\\\\=
(a^{2}+b^{6})(a^4-a^2b^6+b^{12})(a+b^{3})(a^2-ab^3+b^6)(a-b^{3})(a^2+ab^3+b^6)
.\end{array}
