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**Soroban** Hello, 12MED34!

You're right . . . you need the Taylor Series.

Here's the first one . . .

First, we'll crank out the Taylor Series for $\displaystyle (1 + x)^{\frac{1}{3}}$

Formula: .$\displaystyle f(x) \;=\;f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \hdots$

We have: .$\displaystyle \begin{array}{ccccc}f(x)\;= \\ f'(x)\:= \\ f''(x)\,= \\ f'''(x) = \\ \vdots\end{array}$$\displaystyle

\begin{array}{ccccc}(1+x)^{\frac{1}{3}} \\ \frac{1}{3}(1+x)^{\text{-}\frac{2}{3}} \\ \text{-}\frac{2}{9}(1+x)^{\text{-}\frac{5}{3}} \\ \frac{10}{27}(1+x)^{\text{-}\frac{8}{3}} \\ \vdots\end{array}$

Then: .$\displaystyle \begin{array}{ccccc}f(0)\;= \\ f'(0)\:= \\ f''(0)\,= \\ f'''(0)= \\ \vdots\end{array}$$\displaystyle

\begin{array}{ccccc}1 \\ \frac{1}{3} \\ \text{-}\frac{2}{9} \\ \frac{10}{27} \\ \vdots\end{array}$

We have: .$\displaystyle (1+x)^{\frac{1}{3}}\;=\;1 + \frac{1}{3}x -\frac{\frac{2}{9}}{2!}x^2 + \frac{\frac{10}{27}}{3!}x^3 - \hdots$

Hence: .$\displaystyle (1 + x)^{\frac{1}{3}} \;\leq \;1 + \frac{1}{3}x - \frac{1}{9}x^2 + \frac{5}{81}x^2$

Therefore: .$\displaystyle (1 + x)^{\frac{1}{3}} - 1 + \frac{1}{3}x - \frac{1}{9}x^2\;\leq \;\frac{5}{81}x^3$ . *. . . There!*