# How do you use the Binomial Theorem to expand (d-5)^6?

Jul 20, 2017

${d}^{6} - 30 {d}^{5} - 2500 {d}^{3} + 9375 {d}^{2} - 18750 d + 15625$

#### Explanation:

${\left(d + \left(- 5\right)\right)}^{6} = \left(\begin{matrix}6 \\ 0\end{matrix}\right) {d}^{6} {\left(- 5\right)}^{0} + \left(\begin{matrix}6 \\ 1\end{matrix}\right) {d}^{5} {\left(- 5\right)}^{1} +$

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

$\left(\begin{matrix}6 \\ 5\end{matrix}\right) {d}^{1} {\left(- 5\right)}^{5} + \left(\begin{matrix}6 \\ 6\end{matrix}\right) {d}^{0} {\left(- 5\right)}^{6} =$

(6!)/(0!(6-0)!)d^6-5(6!)/(1!(6-1)!)d^5+25(6!)/(2!(6-2)!)d^4+

-125(6!)/(3!(6-3)!)d^3+625(6!)/(4!(6-4)!)d^2-3125(6!)/(5!(6-5)!)d+

15625(6!)/(6!(6-6)!)=

${d}^{6} - 5 \cdot 6 {d}^{5} + 25 \cdot 15 {d}^{4} - 125 \cdot 20 {d}^{3} + 625 \cdot 15 {d}^{2} -$

$3125 \cdot 6 d + 15625 =$

${d}^{6} - 30 {d}^{5} - 2500 {d}^{3} + 9375 {d}^{2} - 18750 d + 15625$