Answer
$\frac{1}{2}$.
Work Step by Step
The given expression is
$=\frac{x^2+16x+64}{2x^2-128}\div \frac{x^2+10x+16}{x^2-6x-16}$
Factor $x^2+16x+64$.
Rewrite the middle term $16x$ as $8x+8x$.
$=x^2+8x+8x+64$
Group terms.
$=(x^2+8x)+(8x+64)$
Factor each term.
$=x(x+8)+8(x+8)$
Factor out $(x+8)$.
$=(x+8)(x+8)$
Factor $2x^2-128$.
$=2(x^2-64)$
$=2(x^2-8^2)$
Use the algebraic identity $a^2-b^2=(a+b)(a-b)$.
$2(x+8)(x-8)$.
Factor $x^2+10x+16$
Rewrite the middle term $10x$ as $8x+2x$.
$=x^2+8x+2x+16$
Group terms.
$=(x^2+8x)+(2x+16)$
Factor each term.
$=x(x+8)+2(x+8)$
Factor out $(x+8)$.
$=(x+8)(x+2)$
Factor $x^2-6x-16$
Rewrite the middle term $-6x$ as $-8x+2x$.
$=x^2-8x+2x-16$
Group terms.
$=(x^2-8x)+(2x-16)$
Factor each term.
$=x(x-8)+2(x-8)$
Factor out $(x-8)$.
$=(x-8)(x+2)$
Substitute the factor into the given expression.
$=\frac{(x+8)(x+8)}{2(x+8)(x-8)}\div \frac{(x+8)(x+2)}{(x-8)(x+2)}$
$=\frac{(x+8)(x+8)}{2(x+8)(x-8)}\times \frac{(x-8)(x+2)}{(x+8)(x+2)}$
Cancel common terms.
$=\frac{1}{2}$.
