How do you use the binomial theorem to expand and simplify the expression #(1/x+2y)^6#?

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Answer 1 · 2018-02-11T22:32:38

#x^-6 + 12yx^-5 + 60y^2x^-4 + 160y^3x^-3 + 240y^4x^-2 + 192y^5x^-1 + 64y^6#

#1#
#1# #1#
#1# #2# #1#
#1# #3 #3 #1#
#1# #4# #6# #4# #1#
#1# #5# #10# #10# #5# #1#
#1# #6# #15# #20# #15# #6# #1# < 6th power

#(1/x + 2y)^6 = 1((1/x)^6) + 6((1/x)^5 2y) + 15((1/x)^4 (2y)^2) + 20((1/x)^3 (2y)^3) + 15((1/x)^2 (2y)^4) + 6(1/x(2y^5)) + 1((2y)^6)#

#(1/x)^6 = 1/x^6 = x^-6#

#6((1/x)^5 2y) = 1/x^5 * 12y = 12yx^-5#

#15((1/x)^4 (2y)^2) = 1/x^4 * 60y^2 = 60y^2x^-4#

#20((1/x)^3 (2y)^3) = 1/x^3 * 160y^3 = 160y^3x^-3#

#15((1/x)^2 (2y)^4) = 1/x^2 * 240y^4 = 240y^4x^-2#

#6(1/x(2y)^5) = 1/x * 192y^5 = 192y^5x^-1#

#(2y)^6 = 64y^6#

total:

#x^-6 + 12yx^-5 + 60y^2x^-4 + 160y^3x^-3 + 240y^4x^-2 + 192y^5x^-1 + 64y^6#

there are no common factors between all of the terms.