${\left(\frac{x}{1-x}\right)}^5$

=  $x^5\sum _{n\=0}^{\infty }\left(\genfrac{}{}{0ex}{}{-5}{n}\right){\left(-x\right)}^n$$$

$\=x^5\left(\sum _{n\=0}^{\infty }\left(1\+\frac{25}{12}n\+\frac{35}{24}{n}^2\+\frac{5}{12}{n}^3\+\frac{1}{24}{n}^4\right)x^n\right)$

$\=x^5\left(1\+5x\+15x^2\+35x^3\+70x^4\+126x^5\+\cdots \right)$

$\=x^5\+5x^6\+15x^7\+35x^8\+70x^9\+126x^{10}\+\cdots$

Obtain Maple's formal power series for ${\left(1-x^3\right)}^{-1\/2}$ 

$$ ${\left(1-x\right)}^{-5}$$\stackrel{\text{formal series}}{\to }$$\sum _{n\=0}^{\mathrm{\∞}}\left(1\+\frac{25}{12}n\+\frac{35}{24}{n}^2\+\frac{5}{12}{n}^3\+\frac{1}{24}{n}^4\right)x^n$

Use this expression to obtain the first few terms of the fulll expansion

$x^5\cdot \sum _{n\=0}^4\left(1\+\frac{25}{12}n\+\frac{35}{24}{n}^2\+\frac{5}{12}{n}^3\+\frac{1}{24}{n}^4\right)x^n$$\=$$x^5\left(70x^4\+35x^3\+15x^2\+5x\+1\right)$$$

$\mathrm{simplify}\left(\mathrm{convert}\left({\left(\frac{x}{1-x}\right)}^5\,\mathrm{FormalPowerSeries}\right)\right)$ = $-1\+\sum _{k\=0}^{\mathrm{\∞}}\frac{1}{24}\left(k^4-10k^3\+35k^2-50k\+24\right)x^k$$$

Obtain the first few terms of the Maclaurin series for ${\left(\frac{x}{1-x}\right)}^5$ 

${\left(\frac{x}{1-x}\right)}^5$$\stackrel{\text{series in x}}{\to }$$210x^{11}\+126x^{10}\+70x^9\+35x^8\+15x^7\+5x^6\+x^5$

Apply the Binomial expansion from first principles

$x^5\sum _{n\=0}^{\infty }\left(\genfrac{}{}{0ex}{}{-5}{n}\right)\cdot {\left(-x\right)}^n$ = $-\frac{x^5}{{\left(-1\+x\right)}^5}$ $$