Answer
$(2t+3)(4t^2-6t+9)$
Work Step by Step
The expressions $
8t^3
$ and $
27
$ are both perfect cubes (the cube root is exact). Hence, $
8t^3+27
$ 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}
(2t)^3+(3)^3
\\\\=
(2t+3)[(2t)^2-2t(3)+(3)^2]
\\\\=
(2t+3)(4t^2-6t+9)
.\end{array}
