# How do I use the the binomial theorem to expand (v - u)^6?

Mar 5, 2018

$\textcolor{b l u e}{{v}^{6} - 6 {v}^{5} u + 15 {v}^{4} {u}^{2} - 20 {v}^{3} {u}^{3} + 15 {v}^{2} {u}^{4} - 6 v {u}^{5} + {u}^{6}}$

#### Explanation:

For a binomial expansion:

${\left(x + y\right)}^{n}$ we have:

$\left(\begin{matrix}n \\ r\end{matrix}\right) {x}^{n - r} {y}^{r}$

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

Where:

((n),(r))=color(white)(0)^n C_(r)=(n!)/(r!(n-r)!)

Beginning with $r = 0$

$\left(\begin{matrix}6 \\ 0\end{matrix}\right) {v}^{6} {\left(- u\right)}^{0} + \left(\begin{matrix}6 \\ 1\end{matrix}\right) {v}^{5} {\left(- u\right)}^{1} + \left(\begin{matrix}6 \\ 2\end{matrix}\right) {v}^{4} {\left(- u\right)}^{2}$

$+ \left(\begin{matrix}6 \\ 3\end{matrix}\right) {v}^{3} {\left(- u\right)}^{3} + \left(\begin{matrix}6 \\ 4\end{matrix}\right) {v}^{2} {\left(- u\right)}^{4} + \left(\begin{matrix}6 \\ 5\end{matrix}\right) {v}^{1} {\left(- u\right)}^{5}$

$\left(\begin{matrix}6 \\ 6\end{matrix}\right) {v}^{0} {\left(- u\right)}^{6}$

Calculating $\left(\begin{matrix}n \\ r\end{matrix}\right)$

$\left(1\right) {v}^{6} {\left(- u\right)}^{0} + \left(6\right) {v}^{5} {\left(- u\right)}^{1} + \left(15\right) {v}^{4} {\left(- u\right)}^{2}$

$+ \left(20\right) {v}^{3} {\left(- u\right)}^{3} + \left(15\right) {v}^{2} {\left(- u\right)}^{4} + \left(6\right) {v}^{1} {\left(- u\right)}^{5}$

$\left(1\right) {v}^{0} {\left(- u\right)}^{6}$

Expand brackets and simplify:

$\textcolor{b l u e}{{v}^{6} - 6 {v}^{5} u + 15 {v}^{4} {u}^{2} - 20 {v}^{3} {u}^{3} + 15 {v}^{2} {u}^{4} - 6 v {u}^{5} + {u}^{6}}$

To make things easier we can use the following:

${\textcolor{w h i t e}{0}}^{n} {C}_{r} = {\textcolor{w h i t e}{0}}^{n} {C}_{n - r}$

And:

${\left(- u\right)}^{n}$ is negative for odd powers and positive for even powers.