Use the binomial series,
$$(1+x{)}^r=\sum _{n=0}^{\mathrm{\infty }}\left(\begin{array}{l}r\\ n\end{array}\right)x^n$$
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 $\boxed{{\displaystyle 1.144}}$.