Problem

Solve for x:
sqrt(x)-3 x^(1/3) = 3 x^(1/6)-9

Subtract 3 x^(1/6)-9 from both sides:
9-3 x^(1/6)-3 x^(1/3)+sqrt(x) = 0

Simplify and substitute y = x^(1/6):
9-3 x^(1/6)-3 x^(1/3)+sqrt(x)  =  9-3 x^(1/6)-3 (x^(1/6))^2+(x^(1/6))^3  =  y^3-3 y^2-3 y+9  =  0:
y^3-3 y^2-3 y+9 = 0

The left hand side factors into a product with two terms:
(y-3) (y^2-3) = 0

Split into two equations:
y-3 = 0 or y^2-3 = 0

Add 3 to both sides:
y = 3 or y^2-3 = 0

Substitute back for y = x^(1/6):
x^(1/6) = 3 or y^2-3 = 0

Raise both sides to the power of six:
x = 729 or y^2-3 = 0

Add 3 to both sides:
x = 729 or y^2 = 3

Take the square root of both sides:
x = 729 or y = sqrt(3) or y = -sqrt(3)

Substitute back for y = x^(1/6):
x = 729 or x^(1/6) = sqrt(3) or y = -sqrt(3)

Raise both sides to the power of six:
x = 729 or x = 27 or y = -sqrt(3)

Substitute back for y = x^(1/6):
x = 729 or x = 27 or x^(1/6) = -sqrt(3)

Raise both sides to the power of six:
Answer: | x = 729                   or             x = 27

Thanks, guest......here's the same problem without using any substitutions :

x^1/2 - 3x^1/3 = 3x^1/6 - 9   .... note....we can  re-write this  as

x^3/6 - 3x^2/6 - 3x^1/6 + 9 = 0     factor directly

(x^2/6 - 3) ( x^1/6 -  3)  = 0

So either

x  ^2/6 - 3  = 0  →  x^1/3 - 3  = 0  → x^1/3  = 3   cube each side   x = 27

Or

x^1/6 - 3 = 0 → x^1/6  = 3     raise each side to the 6th power  x = 3^6 = 3^3 * 3^3  = 27^2  = 729