Answer
$(2r+t)(r^2-rt+19t^2)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
(r+2t)^3+(r-3t)^3
,$ use the factoring of the sum or difference of $2$ cubes.
$\bf{\text{Solution Details:}}$
The expressions $
(r+2t)^3
$ and $
(r-3t)^3
$ are both perfect cubes (the cube root is exact). Hence, $
(r+2t)^3+(r-3t)^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}
[(r+2t)+(r-3t)][(r+2t)^2-(r+2t)(r-3t)+(r-3t)^2]
\\\\=
[2r+t][(r+2t)^2-(r+2t)(r-3t)+(r-3t)^2]
.\end{array}
Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
[2r+t][(r^2+4rt+4t^2)-(r+2t)(r-3t)+(r^2-6rt+9t^2)]
.\end{array}
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel}
[2r+t][(r^2+4rt+4t^2)-(r^2-rt-6t^2)+(r^2-6rt+9t^2)]
.\end{array}
By combining like terms, the expression above is equivalent to
\begin{array}{l}\require{cancel}
[2r+t][r^2+4rt+4t^2-r^2+rt+6t^2+r^2-6rt+9t^2]
\\\\=
[2r+t][(r^2-r^2+r^2)+(4rt+rt-6rt)+(4t^2+6t^2+9t^2)]
\\\\=
[2r+t][r^2-rt+19t^2]
\\\\=
(2r+t)(r^2-rt+19t^2)
.\end{array}
