Question #f64c3

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Question #f64c3

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Answer 1 · 2017-03-08T18:39:33

$2268$ $2268$

Explanation:

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

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

so making

$\frac{1}{3!} \frac{d^{3}}{{d x}^{3}} {(3 y - 1)}^{9} = \frac{1}{3!} 9 \cdot 8 \cdot 7 (3^{3}) = 2268$ $1/(3!)d^3/(dx^3)(3y-1)^9 = 1/(3!)9 cdot 8 cdot 7 (3^3)=2268$

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