Answer
$(2b)(3a^2+b^2)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
(a+b)^3-(a-b)^3
,$ use the factoring of the sum or difference of $2$ cubes.
$\bf{\text{Solution Details:}}$
The expressions $
(a+b)^3
$ and $
(a-b)^3
$ are both perfect cubes (the cube root is exact). Hence, $
(a+b)^3-(a-b)^3
$ 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}
[(a+b)-(a-b)][(a+b)^2+(a+b)(a-b)+(a-b)^2]
\\\\=
[a+b-a+b][(a+b)^2+(a+b)(a-b)+(a+b)^2]
\\\\=
(2b)[(a+b)^2+(a+b)(a-b)+(a-b)^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}
(2b)[(a^2+2ab+b^2)+(a+b)(a-b)+(a^2-2ab+b^2)]
.\end{array}
Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent
\begin{array}{l}\require{cancel}
(2b)[(a^2+2ab+b^2)+(a^2-b^2)+(a^2-2ab+b^2)]
.\end{array}
Removing the grouping symbols and then combining like terms, the expression above is equivalent to
\begin{array}{l}\require{cancel}
(2b)[(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2]
\\\\=
(2b)[(a^2+a^2+a^2)+(2ab-2ab)+(b^2-b^2+b^2)]
\\\\=
(2b)(3a^2+b^2)
.\end{array}
