<s>[INST] How to simplify an expression? I have tried to simplify this expression for quite a long time now but I can't find how to do it.
$$\left(\frac{1}{2+4m}-\frac{1-m}{8m^3+1}:\frac{1-2m}{2m^2-2m+1}\right)\cdot\frac{4m+2}{2m-1}-\frac{1}{1-4m+4m^2}$$
Can someone help me with it?
 [/INST] $\require{cancel}$
Starting with the original expression:
$$\left(\frac{1}{2+4m}-\frac{1-m}{8m^3+1}:\frac{1-2m}{2m^2-2m+1}\right)\cdot\frac{4m+2}{2m-1}-\frac{1}{1-4m+4m^2}$$
Factoring where possible and changing the order of some terms:
$$\left(\frac{1}{2(2m+1)}+\frac{m-1}{(2m+1)(4m^2-2m+1)}:\frac{-(2m-1)}{2m^2-2m+1}\right)\cdot\frac{2(2m+1)}{2m-1}-\frac{1}{(2m-1)^2}$$
Converting division into multiplication by the reciprocal; distributing the outside fraction:
$$\left(\frac{1}{2(2m+1)}\frac{2(2m+1)}{2m-1}+\frac{m-1}{(2m+1)(4m^2-2m+1)}\frac{2m^2-2m+1}{-(2m-1)}\frac{2(2m+1)}{2m-1}\right)-\frac{1}{(2m-1)^2}$$
Some cancellations:
$$\hspace{-0.8mm}\left(\frac{1}{\cancel{2(2m+1)}}\frac{\cancel{2(2m+1)}}{2m-1}+\frac{m-1}{\cancel{(2m+1)}(4m^2-2m+1)}\frac{2m^2-2m+1}{-(2m-1)}\frac{2\cancel{(2m+1)}}{2m-1}\right)-\frac{1}{(2m-1)^2}$$
$$\implies\frac{1}{2m-1}-\frac{2(m-1)}{4m^2-2m+1}\frac{2m^2-2m+1}{(2m-1)^2}-\frac{1}{(2m-1)^2}$$
Finding common denominators:
$$\frac{8m^3-8m^2+4m-1}{(4m^2-2m+1)(2m-1)^2}-\frac{4m^3-8m^2+6m-2}{(4m^2-2m+1)(2m-1)^2}-\frac{4m^2-2m+1}{(4m^2-2m+1)(2m-1)^2}$$
Adding/subtracting:
$$\frac{4m^3-4m^2}{(4m^2-2m+1)(2m-1)^2}$$
Factoring one last time:
$$\frac{4m^2(m-1)}{(4m^2-2m+1)(2m-1)^2}$$
</s>