Hello, kingsolomonsgrave!
$\displaystyle \text{Simplify: }\:\frac{6(x+1)^5(x-3)^2 + 2(x+1)^4(x-3)^3}{(x+1)^{10}}$
I let (x+1)=x and (x-3)=y. . Don't do that!
Use two "new" variables: .$\displaystyle {\color{blue}x + 1 = u,\;x-3 = v}$
and I got $\displaystyle \frac{2x^4y^2(3x+y)}{x^{10}}$
where do I go from here?
You can cancel the x's . . . and back-substitute.
We have: .$\displaystyle \frac{6(x+1)^5(x-3)^2 + 2(x+1)^4(x-3)^3}{(x+1)^{10}}$
Factor: . $\displaystyle \frac{2(x+1)^4(x-3)^2\big[3(x+1) + (x-3)\big]}{(x+1)^{10}} $
Reduce: . $\displaystyle \frac{2(x-3)^2\big[3x+3 + x - 3\big]}{(x+1)^6} $
Simplify: .$\displaystyle \frac{2(x-3)^2\cdot4x}{(x+1)^6} \;=\;\frac{8x(x-3)^2}{(x+1)^6} $
Where did this problem come from?
If it came from an application of the Quotient Rule,
. . something is drastically wrong.