Electric Dipole derivation (Algebra)

[DrummingAtom]

Homework Statement

Show that

$$E = \frac {q}{4\pi\varepsilon(z-\frac{1}{2}d)^2} - \frac{q}{4\pi\varepsilon(z+\frac{1}{2}d)^2}$$

is

$$E = \frac {q}{4\pi\varepsilon z^2} [(1-\frac{d}{2z})^{-2} - (1+\frac{d}{2z})^{-2}]$$

Homework Equations

The Attempt at a Solution

After getting a common denominator and subtracting I got

$$\frac {2qdz}{4\pi\varepsilon(z^4-\frac{1}{2}d^2z^2+\frac{1}{16}d^4)}$$

Stuck completely at this point. I've tried working backwards but that is even more confusing. Any help would be appreciated. Thanks.

 

[N-Gin]

$$E = \frac {q}{4\pi\varepsilon(z-\frac{1}{2}d)^2} - \frac{q}{4\pi\varepsilon(z+\frac{1}{2}d)^2}$$

$$E = \frac {q}{4\pi\varepsilon}[(z-\frac{1}{2}d)^{-2}} - (z+\frac{1}{2}d)^{-2}}]$$

$$E = \frac {q}{4\pi\varepsilon}[(z(1-\frac{d}{2z}))^{-2}} - (z(1+\frac{d}{2z}))^{-2}}]$$

$$E = \frac {q}{4\pi\varepsilon}[z^{-2}(1-\frac{d}{2z})^{-2}} - z^{-2}(1+\frac{d}{2z})^{-2}}]$$

$$E = \frac {q}{4\pi\varepsilon z^2} [(1-\frac{d}{2z})^{-2} - (1+\frac{d}{2z})^{-2}]$$