# How do you use the binomial series to expand (2x^2 – (3/x) ) ^8?

${\left(2 {x}^{2} - \frac{3}{x}\right)}^{8} = 256 {x}^{16} - 3072 {x}^{13} + 16128 {x}^{10} , - 48384 {x}^{7} + 90720 {x}^{4} - 108864 x + 81648 {x}^{- 2} - 34992 {x}^{- 5} + 6561 {x}^{-} 8$

#### Explanation:

Binomial Expansion follows the following general rule:

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

In this question, $a = 2 {x}^{2} , b = - \frac{3}{x}$

I'll use a chart to find each term:

$\left(\begin{matrix}C & a & b & t e r m \\ 1 & 256 {x}^{16} & 1 & 256 {x}^{16} \\ 8 & 128 {x}^{14} & - \frac{3}{x} & - 3072 {x}^{13} \\ 28 & 64 {x}^{12} & \frac{9}{x} ^ 2 & 16128 {x}^{10} \\ 56 & 32 {x}^{10} & - \frac{27}{x} ^ 3 & - 48384 {x}^{7} \\ 70 & 16 {x}^{8} & \frac{81}{x} ^ 4 & 90720 {x}^{4} \\ 56 & 8 {x}^{6} & - \frac{243}{x} ^ 5 & - 108864 x \\ 28 & 4 {x}^{4} & \frac{729}{x} ^ 6 & 81648 {x}^{- 2} \\ 8 & 2 {x}^{2} & - \frac{2187}{x} ^ 7 & - 34992 {x}^{- 5} \\ 1 & 1 & \frac{6561}{x} ^ 8 & 6561 {x}^{-} 8\end{matrix}\right)$

The final answer is the sum of all the terms:

${\left(2 {x}^{2} - \frac{3}{x}\right)}^{8} = 256 {x}^{16} - 3072 {x}^{13} + 16128 {x}^{10} - 48384 {x}^{7} + 90720 {x}^{4} - 108864 x + 81648 {x}^{- 2} - 34992 {x}^{- 5} + 6561 {x}^{-} 8$