Question #f64c3 Precalculus The Binomial Theorem The Binomial Theorem 1 Answer Cesareo R. Mar 8, 2017 #2268# Explanation: Expanding #(3x^3-1)^9=(x(3x^2-1/x))^9# instead, the equivalent term is the one attached to #x^9#. Now making #y=x^3# in #(3y-1)^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# Answer link Related questions What is the binomial theorem? How do I use the binomial theorem to expand #(d-4b)^3#? How do I use the the binomial theorem to expand #(t + w)^4#? How do I use the the binomial theorem to expand #(v - u)^6#? How do I use the binomial theorem to find the constant term? How do you find the coefficient of x^5 in the expansion of (2x+3)(x+1)^8? How do you find the coefficient of x^6 in the expansion of #(2x+3)^10#? How do you use the binomial series to expand #f(x)=1/(sqrt(1+x^2))#? How do you use the binomial series to expand #1 / (1+x)^4#? How do you use the binomial series to expand #f(x)=(1+x)^(1/3 )#? See all questions in The Binomial Theorem Impact of this question 1370 views around the world You can reuse this answer Creative Commons License