# How do you use the binomial theorem to expand and simplify the expression (3a-b)^5?

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

#### Explanation:

Binomial expansion is done using:

${\left(a + b\right)}^{n} = \left({C}_{n , 0}\right) {a}^{n} {b}^{0} + \left({C}_{n , 1}\right) {a}^{n - 1} {b}^{1} + \ldots + \left({C}_{n , n}\right) {a}^{0} {b}^{n}$

And so we have here:

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

${\left(3 a - b\right)}^{5} = \left(1\right) {a}^{5} {b}^{0} + \left(5\right) {a}^{4} {b}^{1} + \left(10\right) {a}^{3} {b}^{2} + \left(10\right) {a}^{2} {b}^{3} + \left(5\right) {a}^{1} {b}^{4} + \left(1\right) {a}^{0} {b}^{5}$

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