2011/099) Find cube root of 9 sqrt 3+11 sqrt 2

because there is sqrt(3) and sqrt(2) in the number the cube root is a combination of them
let it be x sqrt (3) + y sqrt (2)
so (x sqrt (3) + y sqrt (2))^3 = 9 sqrt 3+11 sqrt 2
or 3 x^3 sqrt (3) + 9x^2y sqrt 2 + 6xy^2 sqrt (2) + 2y^3 sqrt 2 = 9 sqrt 3+11 sqrt 2
comparing coefficient of sqrt (3) and sqrt(2) we get
3x^3 + 6x y^2 = 9 ..1
9x^2 y + 2y^3 = 11 .. 2
multiply (1) by 11 and 2nd one by 9 to get
33 x^3 + 66 xy^2 = 81 x^2y + 18y^3
divide both sides by 311 x^3 + 22xy^2 = 27x^2y + 6y^3
or 11x^3 - 27 x^2y + 22 xy^2 - 6y^3 = 0
putting x/y = t we get
11t^3-27 t^2 + 22t- 6 = 0
f(1) = 0 gives t = 1 is a sqrt ( as 11 + 22 = 27+ 6)
so t = 1 or 11t^2-16t + 6 = 0
or x = y is a solution
and
from (1) 9x^3 = 1 or x = 1 and y = 1
so result = sqrt(3) + sqrt (2)