Explanation

(a)
Given,
$\frac{x^{3}}{(2x - 1)(x + 2)(x - 3)} = A + \frac{B}{(2x - 1)} + \frac{C}{(x + 2)} + \frac{D}{(x - 3)}$
Let $f(x) = \frac{x^{3}}{(2x - 1)(x + 2)(x - 3)}$
$= \frac{x^{3}}{2x^{3} - 3x^{2} - 11x + 6}$
Here, the power of $x$ are same in Nr and Dr
$\therefore$ First we divide the numerator by denominator
$2x^{3} - 3x^{2} - 11x + 6\frac{1/2}{x^{3}}$
$x^{3} - \frac{3}{2}x^{2} - \frac{11}{2}x + 3$
$- \ \ \ + \ \ \ \ + \ \ -$
$\frac{3}{2}x^{2} + \frac{11}{2}x - 3$
$\therefore\ \ \ \ \frac{x^{3}}{(2x - 1)(x + 2)(x - 3)} = \frac{1}{2} + \frac{\frac{3}{2}x^{2} + \frac{11}{2}x - 3}{(2x - 1)(x + 2)(x - 3)}$
$\Rightarrow \ \ A = \frac{1}{2}$