Answer
$(x-27)(x^2+3y^2+27x-81y-3xy+729)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
(x-y)^3-(27-y)^3
,$ use the factoring of the sum or difference of $2$ cubes.
$\bf{\text{Solution Details:}}$
The expressions $
(x-y)^3
$ and $
(27-y)^3
$ are both perfect cubes (the cube root is exact). Hence, $
(x-y)^3-(27-y)^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}
[(x-y)-(27-y)][(x-y)^2+(x-y)(27-y)+(27-y)^2]
\\\\=
[x-y-27+y][(x-y)^2+(x-y)(27-y)+(27-y)^2]
\\\\=
[x-27][(x-y)^2+(x-y)(27-y)+(27-y)^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}
[x-27][(x^2-2xy+y^2)+(x-y)(27-y)+(729-54y+y^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}
[x-27][(x^2-2xy+y^2)+(27x-xy-27y+y^2)+(729-54y+y^2)]
.\end{array}
By combining like terms, the expression above is equivalent to
\begin{array}{l}\require{cancel}
[x-27][x^2+(y^2+y^2+y^2)+27x+(-27y-54y)+(-2xy-xy)+729]
\\\\=
(x-27)(x^2+3y^2+27x-81y-3xy+729)
.\end{array}
