Answer
$\left(\dfrac{1}{3}+x\right)\left(\dfrac{1}{9}-\dfrac{1}{3}x+x^2\right)$
Work Step by Step
The expressions $
\dfrac{1}{27}
$ and $
x^3
$ are both perfect cubes (the cube root is exact). Hence, $
\dfrac{1}{27}+x^3
$ 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}
\left(\dfrac{1}{3}\right)^3+(x)^3
\\\\=
\left(\dfrac{1}{3}+x\right)\left[\left(\dfrac{1}{3}\right)^2-\dfrac{1}{3}(x)+(x)^2\right]
\\\\=
\left(\dfrac{1}{3}+x\right)\left(\dfrac{1}{9}-\dfrac{1}{3}x+x^2\right)
.\end{array}
