help

(sqrt(x) - 1)*sqrt(x)*(sqrt(x) + 1)   = 6

                                               x   = 4

(sqrt(4) - 1)*sqrt(4)*(sqrt(4) + 1) = 6

                          (2 - 1)*2*(2 + 1) = 6

                                          1*2*3 = 6

Solve (sqrt(x) - 1)*sqrt(x)*(sqrt(x) + 1) = 6.

(sqrtx-1)*(x+sqrtx)

x sqrtx +x -x-sqrtx = 6

x sqrtx - sqrtx = 6

x^3/2 - x^1/2 = 6    let x = x^1/2  

x^3-x = 6

x(x^2-1) = 6

x(x-1)(x+1) = 6                               Maybe this: assuming x is positive,    factors of 6 are 1 2 3 6  

                                                           then you can deduce x = 2

                                                                 then since we substituted x = x^1/2   ,    x^1/2 =2     or   x = 4

x^1/2 (x^1/2-1)(x^1/2+1) = 6

    x = 4    (from wolframalpha....I did not know how to proceed)

( √x - 1) * √x * *( √x + 1)  = 6

Let  √x =  a  .....so we have

(a - 1)  * a  * ( a + 1)   =  6

(a - 1) (a + 1) a  =  6

(a^2 - 1) a  = 6

a^3  - a   = 6

a^3  - a  - 6   =  0         

By the rational roots theorem, the possible  values for a  = ± [ 1, 2, 3, 6 ]

By inspection     a =   2

Therefore   √x  =  2   ⇒    x  = 4