Algebra Problem how to solve for this

Solve for y: √y+ ∛y=36

the answer here is 729.

Please no backward thingy. I want to know how to eliminate the square root symbols and arrive to 729. Thanks......

y^(1/2) + y^(1/3) = 36
Let u = y^(1/6)
u^3 + u^2 = 36
u^3 + u^2 - 36 = 0
Note u = 3 is a solution. Therefore, u - 3 is a factor of u^3 + u^2 - 36.
..................u^2 + 4u +12
........------------------------------
u - 3 | u^3 + u^2 + 0u - 36
.........u^3 - 3u^2
.........--------------
.................4u^2 + 0u
.................4u^2 -12u
.................-------------
.........................12u - 36
.........................12u - 36
.........................------------
...................................0

(u - 3)(u^2 + 4u + 12) = 0
u - 3 = 0 or u^2 + 4u + 12 = 0
u = 3 or u^2 + 4u + 4 + 8 = 0
u = 3 or (u + 2)^2 = -8
u = 3 or u = two imaginary numbers that can be ignored
3 = y^(1/6)
y = 729

Let u = y^(1/6)

solve for u: u^3 + u^2 = 36

u^3 + u^2 - 36 = 0

u = 3

y = u^6 = 729