standard Taylor series using substitution Find Taylor series using substitution about $0$ for $f(x)=\frac{125}{(5+4x)^3}$ by writing $\frac{125}{(5+4x)^3}=\frac{1}{(1+\frac{4}{5}x)^3}$?
Determine a range of validity for this series.

As Claude did,
$$\frac{1}{1+y}=1-y+y^2-y^3+y^4-y^5+...+(-1)^ny^n...$$ and take second derivative of this expension to get, 
$$ \frac{2}{(1+y)^3}=(\frac{1}{1+y})^{''}=2-6y+12y^2-20y^3...n(n-1)(-1)^ny^{n-2}...$$
$$\frac{1}{(1+y)^3}=1-3y+6y^2-10y^3...{n\choose2}(-1)^{n+1}y^{n-2}... $$
Then replace $y$ by $4x\over5$ then you will get your series.
Since above expension convergent for $|y|<1$ your expension is convergent for $|{4x\over5}|<1$
