# How do you expand (m-a)^5?

Jan 16, 2017

${\left(m - a\right)}^{5} = {m}^{5} - 5 {m}^{4} a + 10 {m}^{3} {a}^{2} - 10 {m}^{2} {a}^{3} + 5 m {a}^{4} - {a}^{5}$

#### Explanation:

${\left(a + b\right)}^{n} = {\sum}_{k = 0}^{n} \left({C}_{n , k}\right) \left({a}^{n - k}\right) \left({b}^{k}\right)$, where

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

In this case

$a = m$

$b = \left(- a\right)$

and

$n = 5$

So,

${\left(m + \left(- a\right)\right)}^{5} = {\sum}_{k = 0}^{5} \left({C}_{n , k}\right) \left({a}^{n - k}\right) \left({b}^{k}\right)$

$= \left({C}_{5 , 0}\right) {m}^{5} {\left(- a\right)}^{0} + \left({C}_{5 , 1}\right) {m}^{5 - 1} {\left(- a\right)}^{1} + \left({C}_{5 , 2}\right) {m}^{5 - 2} {\left(- a\right)}^{2}$

$+ \left({C}_{5 , 3}\right) {m}^{5 - 3} {\left(- a\right)}^{3} + \left({C}_{5 , 4}\right) {m}^{5 - 4} {\left(- a\right)}^{4}$

$+ \left({C}_{5 , 5}\right) {m}^{5 - 5} {\left(- a\right)}^{5}$

$= \left({C}_{5 , 0}\right) {m}^{5} {\left(- a\right)}^{0} + \left({C}_{5 , 1}\right) {m}^{4} {\left(- a\right)}^{1} + \left({C}_{5 , 2}\right) {m}^{3} {\left(- a\right)}^{2}$

$+ \left({C}_{5 , 3}\right) {m}^{2} {\left(- a\right)}^{3} + \left({C}_{5 , 4}\right) {m}^{1} {\left(- a\right)}^{4}$

$+ \left({C}_{5 , 5}\right) {m}^{0} {\left(- a\right)}^{5}$

Since ${x}^{0} = 1$, and ${x}^{1} = x$ we can say

$= \left({C}_{5 , 0}\right) {m}^{5} \left(1\right) + \left({C}_{5 , 1}\right) {m}^{4} \left(- a\right) + \left({C}_{5 , 2}\right) {m}^{3} {\left(- a\right)}^{2}$

$+ \left({C}_{5 , 3}\right) {m}^{2} {\left(- a\right)}^{3} + \left({C}_{5 , 4}\right) m {\left(- a\right)}^{4}$

$+ \left({C}_{5 , 5}\right) \left(1\right) {\left(- a\right)}^{5}$

And since $\left(- x\right) \left(- x\right) = {x}^{2}$, and ${\left({x}^{2}\right)}^{2} = {x}^{4}$ we can also say

$= \left({C}_{5 , 0}\right) {m}^{5} + \left({C}_{5 , 1}\right) {m}^{4} \left(- a\right) + \left({C}_{5 , 2}\right) {m}^{3} {\left(a\right)}^{2}$

$+ \left({C}_{5 , 3}\right) {m}^{2} {\left(- a\right)}^{3} + \left({C}_{5 , 4}\right) m {\left(a\right)}^{4}$

$+ \left({C}_{5 , 5}\right) {\left(- a\right)}^{5}$

Also, on a similar note $\left(- x\right) \left(- x\right) \left(- x\right) = {\left(- x\right)}^{3} = - {x}^{3}$
and ${\left(- x\right)}^{5} = {\left(- x\right)}^{3} {\left(- x\right)}^{2} = {\left(- x\right)}^{3} \left({x}^{2}\right) = - {x}^{5}$

So,

$= \left({C}_{5 , 0}\right) {m}^{5} - \left({C}_{5 , 1}\right) {m}^{4} a + \left({C}_{5 , 2}\right) {m}^{3} {a}^{2}$

$- \left({C}_{5 , 3}\right) {m}^{2} {a}^{3} + \left({C}_{5 , 4}\right) m {a}^{4}$

$- \left({C}_{5 , 5}\right) {a}^{5}$

Since, C_(n,k)=(n!)/(k!(n-k)!), 0! =1 ,and n! = prod_(k=1)^n k =1 times 2 times ... times n

Then

C_(5,0)=(5!)/(0!(5-0)!)=(5!)/((1)(5!))=1

C_(5,1)=(5!)/(1!(5-1)!)=(5!)/((1)(4!))=5

C_(5,2)=(5!)/(2!(5-2)!)=(5!)/((2)(3!))=(4 times 5)/2=20/2=10

C_(5,3)=(5!)/(3!(5-3)!)=(5!)/((3!)(2))=(4 times 5)/2=20/2=10

C_(5,4)=(5!)/(4!(5-4)!)=(5!)/((4!)(1))=5

C_(5,5)=(5!)/(5!(5-5)!)=(5!)/((5!)(0!))=(5!)/((5!)(1))=1

So we plug in and we get

${\left(m - a\right)}^{5} = {m}^{5} - 5 {m}^{4} a + 10 {m}^{3} {a}^{2} - 10 {m}^{2} {a}^{3} + 5 m {a}^{4} - {a}^{5}$