# How do you use the binomial series to expand (a + 3b^3)^5?

Jan 29, 2018

$\textcolor{p u r p \le}{{\left(a + 3 {b}^{3}\right)}^{5} = {a}^{5} + 15 {a}^{4} {b}^{3} + 90 {a}^{3} {b}^{6} + 270 {a}^{2} {b}^{9} + 405 a {b}^{12} + 243 {b}^{15}}$

#### Explanation:

Using Pascal's Triangle to get the coefficients for power 5

## 1 5 10 10 5 1

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

$\textcolor{p u r p \le}{{\left(a + 3 {b}^{3}\right)}^{5} = {a}^{5} + 15 {a}^{4} {b}^{3} + 90 {a}^{3} {b}^{6} + 270 {a}^{2} {b}^{9} + 405 a {b}^{12} + 243 {b}^{15}}$