Question

How do you solve `x^ { \frac { 2} { 3} } - 2x ^ { \frac { 1} { 3} } - 8= 0`?

Answer 1

We use a new variable `u=x^(1/3)` so later we find `x=u^3`

Explanation:

`->u^2-2u-8=0`

This is a normal quadratic equation:
Factorize:

`->(u+2)(u-4)=0`

Two solutions:

(1) `u+2=0->u=-2->x=(-2)^3=-8`

(2) `u-4=0->u=4->x=4^3=64`
graph{x^(2/3)-2x^(1/3)-8 [-42.8, 88.87, -23.9, 42]}