Explanation

(b)
Given expression $\frac{\sqrt{1 + x} + \sqrt[3]{{(1 - x)}^{2}}}{1 + x + \sqrt{1 + x}}$ can be rewritten as
$\frac{{(1 + x)}^{\frac{1}{2}} + {(1 - x)}^{\frac{2}{3}}}{1 + x + {(1 + x)}^{\frac{1}{2}}}$
$= \frac{\left\lbrack 1 + \frac{1}{2}x - \frac{1}{8}x^{2} + ... \right\rbrack + \left\lbrack 1 - \frac{2}{3}x - \frac{1}{9}x^{2} - ... \right\rbrack}{1 + x + \left\lbrack 1 + \frac{1}{2}x - \frac{1}{8}x^{2} + ... \right\rbrack}$
$= \frac{2 - \frac{1}{6}x - \frac{17}{72}x^{2} + ...}{2 + \frac{3}{2}x - \frac{1}{8}x^{2} + ...} = \frac{\left\lbrack 1 - \frac{1}{12}x - \frac{17}{144}x^{2} + ... \right\rbrack}{\left\lbrack 1 + \frac{3}{4}x - \frac{1}{16}x^{2} + ... \right\rbrack}$
$= \left\lbrack 1 - \frac{1}{12}x - \frac{17}{144}x^{2} + ... \right\rbrack\left\lbrack 1 + \frac{3}{4}x - \frac{1}{16}x^{2} + ... \right\rbrack^{- 1}$
$= 1 - \frac{5}{6}x + ...$
$= 1 - \frac{5}{6}x$ (On neglecting the higher powers of $x$)