Problem 20.54

20.54. a)
The potential due to three point charges is given by:
$V = k Q (\frac{1}{x+d} - \frac{2}{x} + \frac{1}{x-d})$
Combining terms yields
$V = kQ(\frac{x(x-d) - 2(x+d)(x-d) + x(x+d)}{x(x+d)(x-d)})$
$V = kQ(\frac{2d^{2}}{x^{3}-xd^{2}})$
or
$V = \frac{2kQd^{2}}{x^{3}-xd^{2}}$

b)
Obviously, if x >> d, then x³ >> xd²
So, the potential reduces to
$V \simeq \frac{2kQd^{2}}{x^{3}}$