# How do you simplify (x-3)^3?

May 5, 2017

${x}^{3} - 9 {x}^{2} + 27 x - 27$

#### Explanation:

$\text{Given " (x+a)(x+b)(x+c)" then expansion is}$

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

$\Rightarrow {\left(x - 3\right)}^{3}$

$= \left(x - 3\right) \left(x - 3\right) \left(x - 3\right)$

$\text{with } a = b = c = - 3$

$\Rightarrow {\left(x - 3\right)}^{3}$

$= {x}^{3} + \left(- 3 - 3 - 3\right) {x}^{2} + \left(9 + 9 + 9\right) x$
color(white)(xx)+(-3)(-3)-3)

$= {x}^{3} - 9 {x}^{2} + 27 x - 27$

An alternate way to work it using binomial expansion

#### Explanation:

An alternate way to do this is to use Binomial Expansion, which uses the general formula of:

${\left(a + b\right)}^{n} = \left({C}_{n , 0}\right) {a}^{n} {b}^{0} + \left({C}_{n , 1}\right) {a}^{n - 1} {b}^{1} + \ldots + \left({C}_{n , n}\right) {a}^{0} {b}^{n}$

So here we have:

• $a = x$
• $b = - 3$
• $n = 3$

$\left(\begin{matrix}C & a & b & \text{term} \\ 1 & {x}^{3} & 1 & {x}^{3} \\ 3 & {x}^{2} & - 3 & - 9 {x}^{2} \\ 3 & x & 9 & 27 x \\ 1 & 1 & - 27 & - 27\end{matrix}\right)$

${x}^{3} - 9 {x}^{2} + 27 x - 27$