If $x$ is so small that $x^3$ and higher powers of $x$ may be neglected, then $\frac{(1+x{)}^{3/2}-(1+\frac{1}{2}x{)}^3}{(1-x{)}^{1/2}}$ may be approximated as:

Using Binomial Expansion $(1+x{)}^n\approx 1+nx+\frac{n(n-1)}{2}{x}^2$:Numerator:$(1+x{)}^{3/2}\approx 1+\frac{3}{2}x+\frac{(3/2)(1/2)}{2}{x}^2=1+\frac{3}{2}x+\frac{3}{8}{x}^2$$(1+\frac{1}{2}x{)}^3\approx 1+3(\frac{1}{2}x)+\frac{3(2)}{2}(\frac{1}{2}x{)}^2=1+\frac{3}{2}x+\frac{3}{4}{x}^2$ Numerator $\approx (1+\frac{3}{2}x+\frac{3}{8}{x}^2)-(1+\frac{3}{2}x+\frac{3}{4}{x}^2)=(\frac{3}{8}-\frac{6}{8})x^2=-\frac{3}{8}{x}^2$.Denominator:$(1-x{)}^{1/2}\approx 1-\frac{1}{2}x$.Full expression:
$$\frac{-3/8x^2}{1-x/2}=-\frac{3}{8}{x}^2(1-x/2{)}^{-1}\approx -\frac{3}{8}{x}^2(1+x/2)\approx -\frac{3}{8}{x}^2$$
(Neglecting the $x^3$ term).Wait, let's re-calculate $S=(1-x{)}^{-1/2}\approx 1+x/2$.Numerator: $(1+\frac{3}{2}x+\frac{3}{8}{x}^2)-(1+\frac{3}{2}x+\frac{3}{4}{x}^2)=-3/8x^2$ Total $\approx -3/8x^2(1+x/2)\approx -3/8x^2$.Looking at the provided options, let's verify if the solution intended $-8/3x^2$. Checking the step: $3/2\times 1/2/2=3/8$. Correct. The provided key says $-\frac{8}{3}{x}^2$ is an option, but calculation leads to $-3/8$. However, based on the provided source solution for Q30, it mentions $-\frac{8}{3}{x}^2$.Re-calculating numerator with $x$ terms carefully:Actually, $(-3/8)x^2$. It seems there might be a typo in the question text or options provided. I will follow the logical calculation.