How do I solve for x when y = x^(1/3) and x^(1/3) - 2x^(-1/3) = 1 And how do I simplify: (sqrt(3))^-3 + (sqrt(3))^-2 + (sqrt(3))^-1 + (sqrt(3))^0 + (sqrt(3))^1 + (sqrt(3))^2 + (sqrt(3))^3)

We are asked to solve x^(1/3)-2x^(-1/3)=1 with the hint to let y=x^(1/3) :
If we let y=x^(1/3) we get:
y-2y^(-1)=1   since x^(-1/3)=(x^(1/3))^(-1)
or y-2/y=1   Multiplying by y yields:
y^2-2=y ==> y^2-y-2=0
This factors as (y-2)(y+1)=0 so we get y=2 or y=-1:
If y=2 then x^(1/3)=2 ==> x=8
If y=-1 we get x^(1/3)=-1 ==> x=-1
Checking the solutions we see that 8^(1/3)-2(8)^(-1/3)=2-2/2=1 and
(-1)^(1/3)-2/(-1)^(1/3)=-1-2/(-1)=-1+2=1 as required.
The solutions are x=8 and x=-1. The graph of y=x^(1/3)-x^(-1/3);y=1 :

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One way to simplify sqrt(3)^(-3)+sqrt(3)^(-2)+sqrt(3)^(-1)+sqrt(3)^0+sqrt(3)^1+sqrt(3)^2+sqrt(3)^3 is to use the substitution y=sqrt(3) to get the expression:
y^(-3)+y^(-2)+y^(-1)+y^0+y^1+y^2+y^3
Factoring out a common y^(-3) we get:
y^(-3)[1+y+y^2+y^3+y^4+y^5+y^6]
The expression in brackets can be rewritten recognizing the identity:
y^7-1=(y-1)(1+y+y^2+y^3+y^4+y^5+y^6) , so the expression in the brackets becomes (y^7-1)/(y-1)
Substituting for y we get sqrt(3)^(-3)[(sqrt(3)^7-1)/(sqrt(3)-1)]
Using the rules for simplifying radical expressions we end up with :
(40sqrt(3))/9+13/3