Simplifying $\frac{b^5-a^5-5b^3a^2+5b^2a^3}{b^4+a^4-2a^2b^2}$ into $\frac{b^3-a^3-2a^2b+2ab^2}{a^2+b^2+2ab}$ I got two algebraic expression, the first should be simplified, with some manipulation I think, to be the second. It's similar to binomial expansion, even though coefficients are a mess.
$$\frac{b^5-a^5-5b^3a^2+5b^2a^3}{b^4+a^4-2a^2b^2} \tag 1$$
$$\frac{b^3-a^3-2a^2b+2ab^2}{a^2+b^2+2ab} \tag 2$$
It's not exactly a problem related to my studies, but something I'm interested in. I remember something similar, when we add/subtract some terms till we got full expansion or so, but I couldn't do anything. Actually, I'm not sure if this problem is correct, so if there's a proof for this problem as invalid one, it's welcome.

The key to this problem is in factoring the denominator. We note the following identity:
$$b^4-2a^2b^2+a^4 = (b^2-a^2)^2 = (b-a)^2(b+a)^2 = (b^2+2ab+a^2)(b^2-2ab+a^2)$$
We see that the denominator splits into the goal denominator and another term. We divide the numerator by $b^2-2ab+a^2$ and get the desired numerator. Here is the polynomial division.
$$(b^2-2ab+a^2)(b^3+2ab^2-2a^2b-a^3)\\
=b^5-2ab^4+a^2b^3+2ab^4-4a^2b^3+2a^3b^2-2a^2b^3+4a^3b^2-2a^4b-a^3b^2-2a^4b-a^5\\
=b^5-5b^3a^2+5b^2a^3-a^5$$
