<s>[INST] 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.
 [/INST] Hint: Notice that substituting $a=b$ gives us numerator as zero for both the expressions. Hence, $(a-b)$ is one of the factors.

 $$1.\ -\frac{(a-b)^3(a^2+3ab+b^2)}{(a^2-b^2)^2}\\ 2.\ -\frac{(a-b)(a^2+3ab+b^2)}{(a+b)^2}$$

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