Is $8x^2-8y^2$ factorable or prime

Question

Edit: From Blitzer Intermediate Algebra 6th edition, page 371

In exercises 23 – 48, factor completely, or state that the polynomial is prime

Edit 2: These set of problems focus on factoring using sum and difference of cubes, although that's not explicitly stated in the instructions. Perhaps within that context $8x^2-8y^2$ is considered prime?

My textbook says $8x^2-8y^2$ is prime, but couldn't $8$ be factored out and then factor using the difference of squares?

$8x^2-8y^2$

$8(x^2-y^2)$

$8(x+y)(x-y)$

Since the textbook claims $8x^2-8y^2$ is prime, why couldn't the polynomial be factored using the above method?

Answer 1

Given the level of math that the book is about, the answer should just be $8x^2-8y^2$ is factorable.