Answer
$\dfrac{1}{2}\left(\dfrac{1}{2}x^{a}+y^{2a}z^{3b} \right)\left(\dfrac{1}{4}x^{2a}-\dfrac{1}{2}x^{a}y^{2a}z^{3b}+y^{4a}z^{6b} \right)$
Work Step by Step
Factoring the $GCF=
\dfrac{1}{2}
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{1}{16}x^{3a}+\dfrac{1}{2}y^{6a}z^{9b}
\\\\=
\dfrac{1}{2}\left(\dfrac{1}{8}x^{3a}+y^{6a}z^{9b} \right)
.\end{array}
The expressions $
\dfrac{1}{8}x^{3a}
$ and $
y^{6a}z^{9b}
$ are both perfect cubes (the cube root is exact). Hence, $
\dfrac{1}{8}x^{3a}+y^{6a}z^{9b}
$ is a $\text{
sum
}$ 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}
\dfrac{1}{2}\left[\left(\dfrac{1}{2}x^{a}\right)^3+\left(y^{2a}z^{3b} \right)^3\right]
\\\\=
\dfrac{1}{2}\left(\dfrac{1}{2}x^{a}+y^{2a}z^{3b} \right)\left(\dfrac{1}{4}x^{2a}-\dfrac{1}{2}x^{a}y^{2a}z^{3b}+y^{4a}z^{6b} \right)
.\end{array}