Explanation

(d)
$\frac{x^{2} + x + 1}{x^{2} + 2x + 1} = 1 - \frac{x}{x^{2} + 2x + 1}$ [on dividing] ...(i)
Now, $\frac{x}{x^{2} + 2x + 1} = \frac{A}{(x + 1)} + \frac{B}{(x + 1)^{2}}$
$\Rightarrow \ \ \ \ \ x = A(x + 1) + B$
On equating the coefficient of $x$ and constant, we get
$A = 1$ and $A + B = 0$
$\Rightarrow \ \ \ A = 1$ and $B = - 1$
From Eq. (i), we get
$\frac{x^{2} + x + 1}{x^{2} + 2x + 1} = 1 - \frac{1}{(x + 1)} + \frac{1}{(x + 1)^{2}}$
$\Rightarrow \ \ \ A + \frac{B}{(x + 1)} + \frac{C}{(x + 1)^{2}} = 1 - \frac{1}{(x + 1)} + \frac{1}{(x + 1)^{2}}$ [given]
$\Rightarrow \ \ A = 1,\ \ \ B = - 1$ and $C = 1$
Now, $A - B = 1 + 1 = 2 = 2C$