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.
 A: 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.
$$(b^2-2ab+a^2)(b^3+2ab^2-2a^2b-a^3)\\
=b^5-2ab^4+a^2b^3+2ab^4-4a^2b^3+2a^3b^2-2a^2b^3+4a^3b^2-2a^4b-a^3b^2-2a^4b-a^5\\
=b^5-5b^3a^2+5b^2a^3-a^5$$
A: The first thing I notice is that $b^4 + a^4 -2a^2b^2$ is $(a^2 - b^2)^2$ and $a^2 + b^2 + 2ab= (a+b)^2$.
I can factor $(a^2-b^2)$ as $(a+b)(a-b)$ so
$(a^2 -b^2)^2 = (a-b)^2(a+b)^2$.
And if we have two fractions equal to each other:
$\frac {????????}{(a-b)^2(a+b)^2} = \frac {!!!!!!!!!}{(a+b)^2}$ that has to mean
$\frac {?????????}{(a-b)^2(a+b)^2} = \frac {!!!!!!!!!}{(a+b)^2}= \frac{!!!!!!!!!\cdot (a-b)^2}{(a-b)^2(a+b)^2}$
So if these are equal we will have to be able to factor $(a-b)^2$ out the first numerator.
So let's try to do that:
$\frac{b^5-a^5-5b^3a^2+5b^2a^3}{b^4+a^4-2a^2b^2}= $
$\frac {(b^5 - a^5)-5(b^3a^2-b^2a^3)}{(b^2 - a^2)^2} = $
$\frac {(b-a)(b^4+b^3a + b^2 a^2+ ba^3 + a^4) - 5b^2a^2(b-a)}{(b-a)^2(b+a)^2}=$ (cancel one of the $(b-a)$ is:
$\frac {b^4 +b^3a + b^2a^2 + ba^3 + a^4 - 5b^2a^2}{(b-a)(b+a)^2} =$
$\frac {b^4 + b^3a -4b^2a^2 + ba^3 + a^4}{(b-a)(b+a)^2}=$ ... now we were told that the fractions were equal so that means $b-a$ factors out the numerator even if it isn't obvious.
$\frac {b^3(b-a)+b^3a + b^3a -4b^2a^2 + ba^3 + a^4}{(b-a)(b+a)^2}=$
$\frac {b^3(b-a)+2 b^3a -4b^2a^2 + ba^3 + a^4}{(b-a)(b+a)^2}=$
$\frac {b^3(b-a)+2 b^2a(b-a)+2b^2a^2 -4b^2a^2 + ba^3 + a^4}{(b-a)(b+a)^2}=$
$\frac {b^3(b-a)+2 b^2a(b-a)-2b^2a^2  + ba^3 + a^4}{(b-a)(b+a)^2}=$
$\frac {b^3(b-a)+2 b^2a(b-a)-2ba^2(b-a)-2ba^3  + ba^3 + a^4}{(b-a)(b+a)^2}=$
$\frac {b^3(b-a)+2 b^2a(b-a)-2ba^2(b-a)- ba^3 + a^4}{(b-a)(b+a)^2}=$
$\frac {b^3(b-a)+2 b^2a(b-a)-2ba^2(b-a)- a^3(b-a)-a^4  + a^4}{(b-a)(b+a)^2}=$
$\frac {b^3(b-a)+2 b^2a(b-a)-2ba^2(b-a)- a^3(b-a)}{(b-a)(b+a)^2}=$
$\frac {b^3 + 2b^2a -2ba^2 - a^3}{(b+a)^2}=$ and that's it, we can rearrange the the terms
$\frac {b^3-a^3-2ba^2 +2b^2a}{b^2 + a^2 + 2ab}$
A: Hint: Notice that substituting $a=b$ gives us numerator as zero for both the expressions. Hence, $(a-b)$ is one of the factors.

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

A: A simpler method would be to define $x=\frac{b}{a}$. And then divide the numerator by $a^5$ and denominator by $a^4$, (and taking a 1/a term separately). Then it's easy to see that $x=1$ is a root of both numerator and denominator. So on dividing numerator and denominator by this factor should solve your question. Also even if you didn't spot the $x=1$ as a common factor, just simply performing long division with numerator and denominator will also solve your question.
