Help in factoring $64x^8+y^8-8x^4-y^4+4x^2y^2$

Question

So a high school student I tutor came to me with this problem and it is the hardest factoring problem I've seen. Anyway, the solution given by Wolfram Alpha is $$(8x^4-4x^2y^2+y^4)(8x^4+4x^2y^2+y^4-1)$$

Upon doing the multiplication we get a bunch of terms cancelling out and we do, in fact, get the original polynomial. Sadly, I cannot see how to get to the factorization.

Answer 1

The original polynomial can be separated into homogeneous parts $$(64x^8+y^8)-(8x^4-4x^2y^2+y^4).$$ If we replace $2x^2$ by $u$ and $y^2$ by $v$ we get $$(4u^4+v^4)-(2u^2-2uv+v^2).$$ Now $$4u^4+v^4=(2u^2+v^2)^2-4u^2v^2 =(2u^2+v^2)^2-(2uv)^2 =(2u^2+v^2+2uv)(2u^2+v^2-2uv).$$ Therefore $$(4u^4+v^8)-(2u^2-2uv+v^2)=(2u^2+2uv+v^2)(2u^2-2uv+v^2) -(2u^2-2uv+v^2)=(2u^2+2uv+v^2-1)(2u^2-2uv+v^2)$$ etc.

Answer 2

$$64x^8+y^8-8x^4-y^4+4x^2y^2 = \color{#952a2a}{(8x^4)^2 +(y^4)^2}-8x^4-y^4+4x^2y^2 \\=\color{#952a2a}{(8x^4 +y^4)^2} - 16x^4y^4-8x^4-y^4+4x^2y^2 \\= (8x^4 +y^4)^2 - \color{blue}{(4x^2y^2)^2}-8x^4-y^4+4x^2y^2 \\=(8x^4 +y^4 -\color{blue}{4x^2y^2})(8x^4 +y^4 + \color{blue}{4x^2y^2})-\color{green}{(8x^4+y^4-4x^2y^2)} \\=(8x^4 +y^4 - 4x^2y^2)(8x^4 +y^4 + 4x^2y^2 - \color{green}{1})$$