Expand #(2x+3)^3# using binomial expansion? Precalculus The Binomial Theorem Pascal's Triangle and Binomial Expansion 1 Answer Shwetank Mauria Apr 28, 2017 #(2x+3)^3=8x^3+12x^2+54x+27# Explanation: Binomial expansion of #(a+b)^n=C_0^na^n+C_1^na^(n-1)b+C_2^na^(n-2)b^2+C_3^na^(n-3)b^3+..++C_n^n)b^n# where #C_r^n=(n(n-1)(n-2).....(n-r+1))/(1.2.3.....r)# and #C_0^n=1# Hence #(2x+3)^3# = #C_0^3(2x)^3+C_1^3(2x)^(3-1)3+C_2^3(2x)^(3-2)3^2+C_3^3(2x)^(3-3)3^3# = #(2x)^3+3/1(2x)^2xx3+(3xx2)/(1xx2)(2x)9+(3xx2xx1)/(1xx2xx3)(2x)^0xx27# = #8x^3+12x^2+54x+27# 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 1804 views around the world You can reuse this answer Creative Commons License