Can you solve x^1/2 - ax^1/3 + bx^1/6 - ab = 0?

Put y = x^(1/6)

x^(1/2) - ax^(1/3) + bx^(1/6) - ab = 0
[x^(1/6)]^3 - [ax^(1/6)]^2 + bx^(1/6) - ab = 0
y^3 - ay^2 + by - ab = 0
y^2(y - a) + b(y - a) = 0
(y - a)(y^2 + b) = 0
y = a or y² = -b (rejected)
x^(1/6) = a
[x^(1/6)]^6 = a^6
x = a^6

You can factor that as
(x^1/6 - a)(x^1/3 + b) = 0
One solution is
x^1/6 = a
x = a^6

Another is
x^1/3 = - b
x = -b^3

x^(1/2) - ax^(1/3) + bx^(1/6) - ab = 0 => let x^(1/6) = u , then: x^(1/3) = u^2, x^(1/2) = u^3:

u^3 - au^2 + bu - ab = 0

u^2(u - a) + b(u - a) = 0

(u^2 + b)(u - a) = 0

u^2 = -b , u = a

x^(1/3) = -b , x^(1/6) = a

x = -b^3 , x = a^6

I hope this helps.

Yes. X*1/2-the total of 1/3 + 1/6 all divided by bx ax a d ab so the answer is x=3