## University Calculus: Early Transcendentals (3rd Edition)

$1-6x+12x^2-8x^3$
The formula to determine the binomial series is: $(1+x)^p=1+\Sigma_{k=1}^\infty \dbinom{p}{k}x^k$ Here, $\dbinom{p}{k}=\dfrac{p(p-1)(p-2).....(p-k+1)}{k!}$ Now, $(1-2x)^{3}=1+3(-2x)+\dfrac{(3)(2)(4x^2)}{2!}+\dfrac{(3)(2)(1)(-8)x^3}{3!}=1-6x+12x^2-8x^3$