How do I factor these expressions (algebra II)

I'm on a homework problem and I'm stuck on how to factor these expressions. I don't remember ever doing them. How do I do them and what are the answers? Thanks :)

1. 8x^3-2x^7

2. x^4-81y^4

3. 8x^3+2x^7

One basic formula you will need for these problems is the difference of two cubes. It states that (a^2-b^2)= (a+b)(a-b)
1. Look at both terms and see what they have in common. They both have a 2 and x^3, so pull these out:
2x^3(4-x^4)
Now, inside the parenthesis, you have a difference of two squares, (4 and x^4 are both squares) so simplify:
****2x^3(2-x^2)(2+x^2)
2. x^4 and 81y^4 are both squares, so use the difference of two squares formula from above:
(x^2+9y^2)(x^2-9y^2)
Notice that one of ther new terms, (x^2-9y^2) is the difference of two cubes, so simplify:
(x+3y)(x-3y)
Now combine this with the other term created in the first step:
*******(x^2+9y^2)(x+3y)(x-3y)
3. Check to see what the terms have in common --> 2 and x^3. Pull out 2x^3:
*****2x^3(4+x^4)
Note: you cannot simplify the addition of two squares- only the difference
hope this helps :)

1) Factor 2x^3

8x^3-2x^7=2x^3(4- x^4)

2x^3(2-x^2)(2+x^2)

2x^3(sqrt(2)-x)(sqrt(2)+x)(2+x^2)

2) This difference between two squares

x^4-81y^4=(x^2 -9y^2)(x^2+9y^2)

=(x-3y)(x+3y)(x^2+9y^2)

3) Same like (1) the only difference it can't be simplified as in (1)

8x^3+2x^7=2x^3(4+ x^4)

8x^3 - 2x^7 = 2[x^3(4 - x^4)] = 2[x^3(2 - x^2)(2 + x^2)]

x^4 - 81y^4 = (x^2 - 9y^2)(x^2 + 9y^2) = (x - 3y(x + 3y)(x^2 + 9y^2)

8x^3 + 2x^7 = 2[x^3(4 + x^4)]