Binomially expand the equation (2+kx)^-3

(2+kx)^-3 = (2^-3)(1+kx/2)^-3 = (2^-3)(1+(-3)(kx/2) + [(-3)(-4)]/2! (kx/2)² + [(-3)(-4)(-5)]/3! (kx/2)³ +... )
= 1⁄8 [1 -(3kx/2) + (12⁄2 k²x²/4) + (60⁄6 k³x³/8) + ...]
= 1⁄8 [1 - (3⁄2 kx) + ( 3⁄2 k²x²) + (5⁄4 k³x³) + ...]
= 1⁄8 - 3⁄16 kx + 3⁄16 k²x² + 5⁄32 k³x³