Factor completely using difference of squares.

x^2-12x+36-64y^2

(x-6)^2-64y^2

(x-6)^2-(8y)^2
(x-6-8y)(x-6+8y)

To factor the expression completely using the difference of squares, we need to rewrite it so that it fits the form of \(a^2 - b^2\).

Let's look at the expression \(x^2-12x+36-64y^2\). Notice that the first three terms form a perfect square trinomial: \((x-6)^2\). Similarly, the last term, -64y^2, is a perfect square with a negative sign: \((8y)^2\).

So, we can rewrite the expression as:
\((x-6)^2 - (8y)^2\)

Now, we have a difference of squares: \(a^2 - b^2\), where \(a = (x-6)\) and \(b = (8y)\).

To factor this completely, we can use the following formula for the difference of squares:
\(a^2 - b^2 = (a+b)(a-b)\)

Applying this formula to our expression, we get:
\((x-6+8y)(x-6-8y)\)

Therefore, the expression \(x^2-12x+36-64y^2\) can be factored completely as \((x-6+8y)(x-6-8y)\).