Proof Verification: Using Taylor series to arrive at a certain inequality If $x>0$ show that $|(1+x)^{1/3} - (1 + \frac{1}{3}x - \frac{1}{9}x^2)| \le \frac{5}{81}x^3$. 
Proof:
Let $f(x) = (1 +x)^{1/3}$
Then, estimating $f(x)$ at the point $x_o = 0$, we have that:
$f'(x) = 1/3(1+x)^{-2/3}, f"(x) = -2/9(1+x)^{-5/3}, f^{(3)}(x)= 10/27 (1+x)^{-8/3}$
Then,
$P_{3} = 1 + \frac{1}{3}x - \frac{1}{9}x^2 + \frac{5}{81}x^3$
And we have that:
$R_3 = \frac{-10}{243} (1+c)^{-11/3}x^4 <0$ for $c \in (0,x)$
Then this implies that:
$(1+x)^{1/3} < 1+ \frac{1}{3}x -\frac{1}{9}x^2 +\frac{5}{81}x^3$
$\implies (1+x)^{1/3}- (1+ \frac{1}{3}x -\frac{1}{9}x^2 ) <\frac{5}{81}x^3$
I don't understand how to transform the above expression into the one that we're required to show. Can anyone explain why it should be $\le$ instead of $<$. Also, how and why do we need the absolute signs?
Please help. Thank you so much.

You've got $(1+x)^{1/3}- (1+ \frac{1}{3}x -\frac{1}{9}x^2 ) <\frac{5}{81}x^3$, but the value on the left side could be a very big negative number, so you actually have to show that the absolute value is smaller than $\frac{5}{81} x^3$.
Working with $P_3, R_3$ is "too much". I suggest to get $P_2,R_2$. Doing that you will get $\leq$ instead of $<$. Feel free to ask me if you don't understand.
