#### Answer

$(4x+y)(16x^2-4xy+y^2-4x+y)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Group the terms of the given expression, $
64x^3+y^3-16x^2+y^2
,$ such that the factored form of the groupings will result to a factor that is common to the entire expression. Then, use the factoring of $2$ cubes and the factoring of the difference of $2$ squares.
$\bf{\text{Solution Details:}}$
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(64x^3+y^3)-(16x^2-y^2)
.\end{array}
The expressions $
64x^3
$ and $
y^3
$ are both perfect cubes (the cube root is exact). Hence, $
64x^3+y^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}
[(4x)^3+(y)^3)]-(16x^2-y^2)
\\\\=
[(4x+y)(16x^2-4xy+y^2)]-(16x^2-y^2)
\\\\=
(4x+y)(16x^2-4xy+y^2)-(16x^2-y^2)
.\end{array}
The expressions $
16x^2
$ and $
y^2
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
16x^2-y^2
,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(4x+y)(16x^2-4xy+y^2)-[(4x)^2-(y)^2]
\\\\=
(4x+y)(16x^2-4xy+y^2)-[(4x+y)(4x-y)]
\\\\=
(4x+y)(16x^2-4xy+y^2)-(4x+y)(4x-y)
.\end{array}
Factoring the $GCF=
(4x+y)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(4x+y)[(16x^2-4xy+y^2)-(4x-y)]
\\\\
(4x+y)[16x^2-4xy+y^2-4x+y]
\\\\
(4x+y)(16x^2-4xy+y^2-4x+y)
.\end{array}