# Question 78619

Aug 24, 2017

${\left(x - 7\right)}^{3} = {x}^{3} - 21 {x}^{2} + 147 x - 343$

#### Explanation:

For low powers you can make use of Pascal's Triangle but I'll do it more generally here with the binomial theorem:

${\left(x + y\right)}^{n} = {\sum}_{k = 0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) {x}^{n - k} {y}^{k}$

where ((n),(k)) = (n!)/((n-k)!k!

Hence, we have

${\left(x - 7\right)}^{3} = {\sum}_{k = 0}^{3} \left(\begin{matrix}3 \\ k\end{matrix}\right) {x}^{3 - k} {y}^{k}$

$= \left(\begin{matrix}3 \\ 0\end{matrix}\right) {x}^{3} + \left(\begin{matrix}3 \\ 1\end{matrix}\right) {x}^{2} \left(- 7\right) + \left(\begin{matrix}3 \\ 2\end{matrix}\right) x {\left(- 7\right)}^{2} + \left(\begin{matrix}3 \\ 3\end{matrix}\right) {\left(- 7\right)}^{3}$

we evaluate this, noting that 0! = 1#

$= {x}^{3} - 21 {x}^{2} + 147 x - 343$

Aug 24, 2017

${x}^{3} - 21 {x}^{2} + 147 x - 343$

#### Explanation:

$\text{to expand "(x+a)^3" in general}$

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

$\text{here } a = - 7$

$\Rightarrow {\left(x - 7\right)}^{3}$

$= {x}^{3} + \left(- 7 - 7 - 7\right) {x}^{2} + \left(49 + 49 + 49\right) x + {\left(- 7\right)}^{3}$

$= {x}^{3} - 21 {x}^{2} + 147 x - 343$