#1:

Take the second part, its easier.

(x-4)²(x+4)²= [(x-4)(x+4)]², much easier that way - you'll get a binomial squared.

For the first part, i'd do

[(x+2)(x-1)]²=(x²+x-2)²

then i'd just crank a trinomial squared, it's like foil except you have more combinations (9 combinations).

2. Find the common denominator: should be

a²b+ab²...

gives you

(a+b)(ab+b²)-(a-b)(a)-(2a+b)(a²+b)

all over

(a²b+ab²)

simplifying the numerator gives you the answer in simplified form, you might be able to factor out and cancel the a's and b's after that.

For the restrictions, you look at the denominators for the original problem: we can't divide by 0 (impossible!) so none of the denominators can equal 0.

... In other words (<> means "does not equal"),

a<>0, ab+b^2 <> 0, and b <> 0.

solving the second equation gives

b(a+b) <> 0,

a+b <>0

a<>-b