# Question f64c3

Mar 8, 2017

$2268$

#### Explanation:

Expanding ${\left(3 {x}^{3} - 1\right)}^{9} = {\left(x \left(3 {x}^{2} - \frac{1}{x}\right)\right)}^{9}$ instead, the equivalent term is the one attached to ${x}^{9}$.

Now making $y = {x}^{3}$ in ${\left(3 y - 1\right)}^{9}$ will be equivalently the constant associated to the ${y}^{3}$ term

so making

1/(3!)d^3/(dx^3)(3y-1)^9 = 1/(3!)9 cdot 8 cdot 7 (3^3)=2268#