How do you use the Binomial Theorem to expand # (1/X + X)^6#?

1 Answer
Nov 7, 2016

Answer:

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

Explanation:

Binomial expansion of #(x+a)^6# is

#x^6+^6C_1x^5a+^6C_2x^4a^2+^6C_3x^3a^3+^6C_4x^2a^4+^6C_5xa^5+^6C_6a^6#

And hence binomial expansion of #(1/x+x)^6# or #(x+1/x)^6# is

#x^6+^6C_1x^5(1/x)+^6C_2x^4(1/x)^2+^6C_3x^3(1/x)^3+^6C_4x^2(1/x)^4+^6C_5x(1/x)^5+^6C_6(1/x)^6#

or #x^6+6/1xx x^4+(6xx5)/(1xx2)xx x^2+(6xx5xx4)/(1xx2xx3)+(6xx5xx4xx3)/(1xx2xx3xx4)xx 1/x^4+(6xx5xx4xx3xx2)/(1xx2xx3xx4xx5)xx 1/x^6#

or #x^6+6x^4+15x^2+20+15/x^2+6/x^4+1/x^6#

or #1/x^6+6/x^4+15/x^2+20+15x^2+6x^4+1/x^6#