#### Answer

(y-8)(y-8)

#### Work Step by Step

$y^{2}$-16y+64
Using the rule of
$a^{2}$-2ab+$b^{2}$= $(a-b)^{2}$
In this case, the a= y and b= 8 as substituting these gives us the original polynomial
$a^{2}$-2ab+$b^{2}$
$y^{2}$-2(y)(8)+$8^{2}$
$y^{2}$-16y+64
Thus:
$y^{2}$-16y+64
= $(y-8)^{2}$
= (y-8)(y-8)