# How do you expand (r+s)^8?

Dec 27, 2016

${\left(r + s\right)}^{8} = {r}^{8} + 8 {r}^{7} s + 28 {r}^{6} {s}^{2} + 56 {r}^{5} {s}^{3} + 70 {r}^{4} {s}^{4} + 56 {r}^{3} {s}^{5} + 28 {r}^{2} {s}^{6} + 8 r {s}^{7} + {s}^{8}$

#### Explanation:

${\left(r + s\right)}^{8} = {\sum}_{k = 0}^{8} \left(\begin{matrix}8 \\ k\end{matrix}\right) {r}^{8 - k} {s}^{k}$

where ((8),(k)) = (8!)/((8-k)! k!)

Rather than calculate these binomial coefficients explicitly, we can read them from the $9$th row of Pascal's triangle (i.e. the row beginning $1 , 8 , \ldots$)

So:

${\left(r + s\right)}^{8} = {r}^{8} + 8 {r}^{7} s + 28 {r}^{6} {s}^{2} + 56 {r}^{5} {s}^{3} + 70 {r}^{4} {s}^{4} + 56 {r}^{3} {s}^{5} + 28 {r}^{2} {s}^{6} + 8 r {s}^{7} + {s}^{8}$