# 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 `

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