Algebra Problem: 64x^4 - 4y^4?

You can't "solve" it because it isn't an equation. All you can do is factor it.

64x^4-4y^4
first factor out 4
4(16x^4-y^4)
This is the difference of two squares. a²-b²=(a+b)(a-b)
In this case a=4x² and b=y²
4(4x²+y²)(4x²-y²)
again we have the difference of squares so we can factor some more
4(4x²+y²)(2x+y)(2x-y)

Assuming it's set equal to zero:
add 4y^4 to both sides
64x^4=4y^4
divide both sides by 4
16x^4=y^4
take the 4th root of both sides
+/-(2x)=y
then plug in both values of y in the original equation to solve in terms of x

64x^4 - 4y^4=
(8x^2)^2-(2y^2)^2=
(8x^2-2y^2)(8x^2+2y^2)=
4[(2x)^2-y^2][4x^2+y^2]=
4(2x-y)(2x+y)(4x^2+y^2)

a^2 - b^2 = (a + b)(a - b)

64x^4 - 4y^4
= 4(16x^4 - y^4)
= 4(4x^2 + y^2)(4x^2 - y^2)
= 4(4x^2 + y^2)(2x + y)(2x - y)

4 (16x^4 - y^4)
4 (4x² - y²)(4x² + y²)
4 (2x - y)(2x + y) (4x² + y²)

That's not a problem. That's just a polynomial. That's like asking someone to solve 3.