Answer
$(x^{2a}-y^{b})(x^{4a}+x^{2a}y^{b}+y^{2b})$
Work Step by Step
The expressions $
x^{6a}
$ and $
y^{3b}
$ are both perfect cubes (the cube root is exact). Hence, $
x^{6a}-y^{3b}
$ is a $\text{
difference
}$ of $2$ cubes. 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^{2a})^3-(y^{b})^3
\\\\=
(x^{2a}-y^{b})[(x^{2a})^2+x^{2a}(y^{b})+(y^{b})^2]
\\\\=
(x^{2a}-y^{b})(x^{4a}+x^{2a}y^{b}+y^{2b})
.\end{array}
