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.
Hint

Start with Taylor expansion $$\frac{1}{1+y}=1-y+y^2-y^3+y^4-y^5+O\left(y^6\right)$$ Rise to the third power to obtain  $$\frac{1}{(1+y)^3}=1-3 y+6 y^2-10 y^3+15 y^4-21 y^5+O\left(y^6\right)$$ Now, replace $y$ by $\frac{4 x}{5}$ to get your result.

I am sure that you can take from here.
you have $(1+4/5x)^{-3}$=1-12/5x+64/25$x^2$.....
is the required exxpansion about x=0. for range, its true for |4/5x|&lt;1