How do you find the x-int's of the function: 1/3x^3 - 9x + 2

[Math_Junkie]

y = 1/3x^3 - 9x + 2

x-int, let y = 0
1/3x^3 - 9x + 2 = 0
x^3 - 27x + 6 = 0

This is where I get stuck, how do you factor that and solve for x?
Thanks in advance for any help.

 

[Deleted member 4993]

y = 1/3x^3 - 9x + 2

x-int, let y = 0
1/3x^3 - 9x + 2 = 0
x^3 - 27x + 6 = 0

This is where I get stuck, how do you factor that and solve for x?
Thanks in advance for any help.

If this is a high school problem, I would apply rational root theorem first and check if 1,2,3 or 6 is one of the roots or not.

If neither of those apply - then I would try graphical solution.