# How do you use the binomial series to expand (1 + 4x)^-3?

${\left(1 + 4 x\right)}^{-} 3 = \frac{1}{1 + 4 x} ^ 3 = \frac{1}{1 + 12 x + 48 {x}^{2} + 64 {x}^{3}}$

#### Explanation:

From the binomial theorem formula

${\left(a + b\right)}^{3} = {a}^{3} + 3 {a}^{2} b + 3 a {b}^{2} + {b}^{3}$

We let $a = 1$ and $b = 4 x$

and

${\left(1 + 4 x\right)}^{3} = {\left(1\right)}^{3} + 3 {\left(1\right)}^{2} \left(4 x\right) + 3 \left(1\right) {\left(4 x\right)}^{2} + {\left(4 x\right)}^{3}$

${\left(1 + 4 x\right)}^{3} = 1 + 12 x + 48 {x}^{2} + 64 {x}^{3}$

This will be the denominator of $\frac{1}{1 + 4 x} ^ 3$

God bless....I hope the explanation is useful.