Use the binomial series,
\[
(1+x)^{r}=\sum_{n=0}^{\infty}\left(\begin{array}{l}
r \\
n
\end{array}\right) x^{n}
\]
to find a $4^{\text {th }}$ order Maclaurin polynomial in order to estimate $\left(\frac{2}{3}\right)^{-\frac{1}{3}}$. Round the answer to three decimal places.

Step 1 ：We are asked to find the 4th order Maclaurin polynomial for the function \((\frac{2}{3})^{-\frac{1}{3}}\).

Step 2 ：We can rewrite this function as \((1 - \frac{1}{3})^{-\frac{1}{3}}\). This is in the form of \((1+x)^r\) where \(x = -\frac{1}{3}\) and \(r = -\frac{1}{3}\).

Step 3 ：The 4th order Maclaurin polynomial is given by the sum of the first 5 terms of the binomial series. We can calculate these terms using the binomial coefficient formula, \(\left(\begin{array}{l} r \\ n \end{array}\right) = \frac{r(r-1)(r-2)...(r-n+1)}{n!}\), and the power of x, \(x^n\).

Step 4 ：Substituting the values of x and r into the binomial series, we get a polynomial value of approximately 1.1439821165472746.

Step 5 ：Final Answer: The 4th order Maclaurin polynomial for the function \((\frac{2}{3})^{-\frac{1}{3}}\) is approximately \(\boxed{1.144}\).