# How do you use the Binomial Theorem to expand (d-3b)^3?

Sep 4, 2017

The answer is $= {d}^{3} - 9 {d}^{2} b + 27 {\mathrm{db}}^{2} - 27 {d}^{3}$

#### Explanation:

${\left(x + y\right)}^{n} = \left(\begin{matrix}n \\ 0\end{matrix}\right) {x}^{n} + \left(\begin{matrix}n \\ 1\end{matrix}\right) {x}^{n - 1} y + \left(\begin{matrix}n \\ 2\end{matrix}\right) {x}^{n - 2} {y}^{2} + \ldots \ldots . . + \left(\begin{matrix}n \\ n\end{matrix}\right) {y}^{n}$

$\forall n \in \mathbb{N}$ and $x , y \in \mathbb{R}$

When $n = 3$

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

Therefore,

${\left(d - 3 b\right)}^{3} = {d}^{3} + 3 {d}^{2} \cdot \left(- 3 b\right) + 3 d \cdot {\left(- 3 d\right)}^{2} + {\left(- 3 d\right)}^{3}$

$= {d}^{3} - 9 {d}^{2} b + 27 {\mathrm{db}}^{2} - 27 {d}^{3}$