factor:

1) (x+y)^2 - (a+b)^2
= [(x+y) - (a+b)] [(x+y) + (a+b)]
= [x+y-a-b] [x+y+a+b]

2) 6x^4 - 6
= 6(x^4 - 1)
= 6(x^2-1)(x^2+1)

Is this right?

1. x^2 +2xy +y^2 -a^2-2ab-B^2

2. 6(x^4-1) = 6(x^2-1)(x^2 +1)

= 6(x-1)(x+1)(x^2+1)

#2 is ok, but not yet complete:

x^2-1 = (x-1)(x+1)

Yes, your factorizations are correct! To explain how you arrived at each factorization, I will break it down step by step:

1) (x+y)^2 - (a+b)^2:
To factor this expression, you can use the difference of squares formula: a^2 - b^2 = (a+b)(a-b). In this case, a = (x+y) and b = (a+b). So using the difference of squares formula, we get:
(x+y)^2 - (a+b)^2 = [(x+y) + (a+b)][(x+y) - (a+b)].
Simplifying further, we have:
(x+y)^2 - (a+b)^2 = [x+y-a-b][x+y+a+b].

2) 6x^4 - 6:
To factor this expression, we can first factor out the greatest common factor, which is 6:
6x^4 - 6 = 6(x^4 - 1).
Now, we have a difference of squares: x^2 - 1 = (x-1)(x+1). So we can write:
6(x^4 - 1) = 6(x^2 - 1)(x^2 + 1).
This is the fully factored expression.

Great job on your factorizations!