How do you expand #(x^3-1/x^2)^3#? Precalculus The Binomial Theorem Pascal's Triangle and Binomial Expansion 1 Answer sjc Jun 12, 2018 #x^9-3x^4+3/x-1/x^6# Explanation: now #(a+b)^3=a^3+3a^2b+3ab^2+b^3--(1)# #(x^3-1/x^2)^3# #a=x^3, b=-1/x^2# substituting into #(1)# #=(x^3)^3+3(x^3)^2(-1/x^2)+3x^3(-1/x^2)^2+(-1/x^2)^3# now simplify #=x^9-3x^6/x^2+3x^3/x^4-1/x^6# #=x^9-3x^4+3/x-1/x^6# Answer link Related questions What is Pascal's triangle? How do I find the #n#th row of Pascal's triangle? How does Pascal's triangle relate to binomial expansion? How do I find a coefficient using Pascal's triangle? How do I use Pascal's triangle to expand #(2x + y)^4#? How do I use Pascal's triangle to expand #(3a + b)^4#? How do I use Pascal's triangle to expand #(x + 2)^5#? How do I use Pascal's triangle to expand #(x - 1)^5#? How do I use Pascal's triangle to expand a binomial? How do I use Pascal's triangle to expand the binomial #(a-b)^6#? See all questions in Pascal's Triangle and Binomial Expansion Impact of this question 2273 views around the world You can reuse this answer Creative Commons License