Please understand that #(x^(1/6))^6 = x^(6/6) = x^1 = x# and the denominator will be rational.
Let #u = x^(1/6)#, then #x = u^6
The denominator that we want can be written in terms of u as:
#-u^6+1#
The denominator that we have can be written in terms of u as:
#u + 1#
We can find #P(u)# by dividing the denominator that we want by the denominator that we have (please observe that I have filled in the missing terms with 0s):
#color(white)( (u+1)/color(black)(u+1))(-u^5+u^4-u^3+u^2-u+1color(white)(..+1))/(")" color(white)(x)-u^6+0u^5+0u^4+0u^3+0u^2+0u+1)#
#P(u) = -u^5+u^4-u^3+u^2-u+1#
Reverse the substitution:
#P(x) = -x^(5/6)+x^(4/6)-x^(3/6)+x^(2/6)-x^(1/6)+1#
We multiply the given expression by 1 in the form of #(P(x))/(P(x))#
#1/(1+x^(1/6))(-x^(5/6)+x^(4/6)-x^(3/6)+x^(2/6)-x^(1/6)+1)/(-x^(5/6)+x^(4/6)-x^(3/6)+x^(2/6)-x^(1/6)+1) =#
#(-x^(5/6)+x^(4/6)-x^(3/6)+x^(2/6)-x^(1/6)+1)/(1-x)#
This has rationalized the denominator.
