# How do you find the product (5g+2h)^3?

$255 {g}^{3} + 150 {g}^{2} h + 60 g {h}^{2} + 8 {h}^{3}$

#### Explanation:

We can work this in a couple of ways.

• One way is to manually multiply the terms the "long way". I'll decline to work this problem that way here.

• The other way is to use Binomial Expansion. That's the way I'll work the problem.

The general form for binomial expansion is:

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

In this case, $a = 5 g , b = 2 h , n = 3$

${\left(5 g + 2 h\right)}^{3}$

$= \left({C}_{3 , 0}\right) {\left(5 g\right)}^{3} {\left(2 h\right)}^{0} + \left({C}_{3 , 1}\right) {\left(5 g\right)}^{2} {\left(2 h\right)}^{1} + \left({C}_{3 , 2}\right) {\left(5 g\right)}^{1} {\left(2 h\right)}^{2} + \left({C}_{3 , 3}\right) {\left(5 g\right)}^{0} {\left(2 h\right)}^{3}$

$= \left(1\right) \left(255\right) \left({g}^{3}\right) \left(1\right) + \left(3\right) \left(25\right) \left({g}^{2}\right) \left(2 h\right) + \left(3\right) \left(5 g\right) \left(4\right) \left({h}^{2}\right) + \left(1\right) \left(1\right) \left(8\right) \left({h}^{3}\right)$

$= 255 {g}^{3} + 150 {g}^{2} h + 60 g {h}^{2} + 8 {h}^{3}$