How to do you solve...x the the power of 2/3 minus 5x to the power of 1/3 minus 6 equals zero..What is the ?

x^(2/3) - 5 x^(1/3) - 6 = 0

Let y = x^(1/3)

y² - 5 y - 6 = 0
(y - 6)(y + 1 ) = 0
y = 6 , y = - 1

x^(1/3) = 6 , x^(1/3) = - 1
x = 216 , x = - 1

x^(2/3) - 5x^(1/3) - 6 = 0
x^(2/3) + x^(1/3) - 6x^(1/3) - 6 = 0
[x^(2/3) + x^(1/3)] - [6x^(1/3) + 6] = 0
x^(1/3)[x^(1/3) + 1] - 6[x^(1/3) + 1] = 0
[x^(1/3) + 1][x^(1/3) - 6] = 0

x^(1/3) + 1 = 0
x^(1/3) = -1
x = -1^(3/1)
x = -1

x^(1/3) - 6 = 0
x^(1/3) = 6
x = 6^(3/1)
x = 216

∴ x = -1 , 216

x^(2/3) - 5x^(1/3) - 6 = 0 , let u = x^(1/3)

then we get:

u^2 - 5u - 6 = 0 , we can solve this equation

u1 = -1
u2 = 6

u1 = x^(1/3)
u1^3 = x
-1^3 = x
-1 = x

u2 = x^(1/3)
u2^3 = x
6^3 = x
216 = x

x is -1 and 216

Let y = x^(1/3)

This means

y^3 = x

Your equation becomes

y^2 - 5y - 6

this factors to:

(y - 6)(y + 1)

the zeros are

y = 6

y = -1

Since y^3 = x

x = 216

and

x= -1