Answer
$(y+z+4)(y^2+2yz+z^2-4y-4z+16)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
(y+z)^3+64
,$ use the factoring of the sum/difference of $2$ cubes.
$\bf{\text{Solution Details:}}$
Using the factoring of the sum/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}
[(y+z)+4][(y+z)^2-(y+z)(4)+(4)^2]
\\\\=
(y+z+4)[(y+z)^2-4y-4z+16]
.\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}
(y+z+4)[(y^2+2yz+z^2)-4y-4z+16]
\\\\=
(y+z+4)(y^2+2yz+z^2-4y-4z+16)
.\end{array}