#(y-4)^3# is simplified.
However, if you want to expand it using binomial expansion, you can use Pascal's triangle:
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#
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: simplify #(y-4)^3#
Let #a = y " and " b = -4 ; " " n =3#
From Pascal's triangle: # ""_3C_0 = ""_3C_3 = 1; ""_3C_1 = ""_3C_2 = 3 #
#(y-4)^3 =#
# ""_3C_0 y^3 (-4)^0 + ""_3C_1 y^2 (-4)^1 + ""_3C_2 y^1 (-4)^2 + ""_3C_3 y^0 (-4)^3 = #
#y^3 + 3y^2(-4) + 3y(16) -64 = #
#y^3 -12y^2 +48y - 64#
