Please understand that (x^(1/6))^6 = x^(6/6) = x^1 = x(x16)6=x66=x1=x and the denominator will be rational.
Let u = x^(1/6)u=x16, then #x = u^6
The denominator that we want can be written in terms of u as:
-u^6+1−u6+1
The denominator that we have can be written in terms of u as:
u + 1u+1
We can find P(u)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)u+1u+1−u5+u4−u3+u2−u+1..+1)x−u6+0u5+0u4+0u3+0u2+0u+1
P(u) = -u^5+u^4-u^3+u^2-u+1P(u)=−u5+u4−u3+u2−u+1
Reverse the substitution:
P(x) = -x^(5/6)+x^(4/6)-x^(3/6)+x^(2/6)-x^(1/6)+1P(x)=−x56+x46−x36+x26−x16+1
We multiply the given expression by 1 in the form of (P(x))/(P(x))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) =11+x16−x56+x46−x36+x26−x16+1−x56+x46−x36+x26−x16+1=
(-x^(5/6)+x^(4/6)-x^(3/6)+x^(2/6)-x^(1/6)+1)/(1-x)−x56+x46−x36+x26−x16+11−x
This has rationalized the denominator.
