If $x>0$ Show that $\lvert (1+x)^{(1/3)} - (1+\frac{x}{3} -\frac{x^2}{9}) \rvert \le (\frac {5}{81})x^3$. Use this inequality to approximate $1.2^{1/3}$ & $2^{1/3}$.
That is the actual problem. I'm assuming that I would use induction to start this proof (correct me if I am wrong). I already proved the base case for when $x = 1$. My issue is showing that, this is true for every $x \gt 0$ in the inductive step. I attempted to do this algebraically but I'm not having any luck. I also would like some help on the approximations as I have no idea how to incorporate this. Any help on this would be appreciated.
Hint:
Assuming that $n$ should be $x$ instead, you want to use that
$\displaystyle\left|(1+x)^{1/3}-(1+\frac{x}{3}-\frac{x^2}{9})\right|=|R_{2}(x)|=\left|\frac{f^{(3)}(c)}{3!}(x-0)^3\right|$,
where $0<c<x$ and $f^{(3)}(x)=\frac{10}{27}(1+x)^{-\frac{8}{3}}$.