Use the Binomial Theorem and/or Pascal's triangle. Pascal's triangle give the combination values.
Binomial Theorem: #(a + b)^n = #
# ""_nC_0 a^n b^0 + _nC_1 a^(n-1) b^1 + _nC_2a^(n-2)b^2 + ... + _nCn a^0 b^n#
where # ""_nC_0 = (n!)/((n-0!)(0!)) = 1; " " _nC_1 = (n!)/((n-1!)(1!))#
# ""_nC_2 = (n!)/((n-2!)(2!)); " "_nC_n = (n!)/((n-n!)(n!)) = 1#
Given: # (y^3 - 2x^4)^5 #
Using the Binomial Theorem:
Let #a = y^3 " and " b = -2x^4 ; " " n =5#
# (y^3 - 2x^4)^5 = ""_5C_0 (y^3)^5 (-2x^4)^0 + _5C_1 (y^3)^4 (-2x^4)^1 + _5C_2 (y^3)^3 (-2x^4)^2 + _5C_3 (y^3)^2 (-2x^4)^3 + _5C_4 (y^3)^1 (-2x^4)^4 + _5C_5 (y^3)^0 (-2x^4)^5#
Calculate the combinations:
# ""_5C_0 ="" _5C_5 = 1#
#""_5C_1 = (5!)/((5-1!)(1!)) = ""_5C_4 = (5!)/((5-4!)(4!)) = 5#
#""_5C_2 = (5!)/((5-2!)(2!)) = ""_5C_3 = (5!)/((5-3!)(3!)) = 10#
Substitute in the values and use the exponent rules
#(2x^my^n)^c = 2^c x^(m*c) y^(m*c) " "and (2x^m)^0 = 1#
# (y^3 - 2x^4)^5 = y^15 +5y^12 (-2x^4) + 10 y^9 (-2)^2x^8 + 10y^6 (-2)^3 x^12 + 5y^3 (-2)^4x^16 +(-2)^5x^20#
#(y^3 - 2x^4)^5 = y^15 -10x^4y^12 + 40x^8y^9 - 80x^12y^6 + 80x^16y^3 - 32x^20#
Pascal's triangle gives the combinations:
#" "1# #" "(a + b)^0#
#" "1 " " 1"# #" "(a + b)^1#
#" "1 " "2" " 1"# #" "(a + b)^2#
#" "1 " " 3 " " 3 " "1 # #" "(a + b)^3#
#" "1 " " 4 " " 6 " " 4 " " 1# #" "(a + b)^4#
#" "1 " " 5 " " 10 " " 10 " "5 " " 1# #" "(a + b)^5#
