How to solve for x, if x^3-x^2+(4/27)=0?

x³ - x² + (4/27) = 0
27x³ - 27x² + 4 = 0

Put f(x) = 27x³ - 27x² + 4
f(-1/3) = 27(-1/3)³ - 27(-1/3)² + 4 = 0
By Factor Theorem, (3x + 1) is one of the factors of (27x³ - 27x² + 4).

27x³ - 27x² + 4 = 0
(27x³ + 9x²) - 9x² - 27x² + 4 = 0
(27x³ + 9x²) + (-36x² - 12x) + 12x + 4 = 0
(27x³ + 9x²) + (-36x² - 12x²) + (12x² + 4) = 0
9x²(3x + 1) - 12x(3x + 1) + 4(3x + 1) = 0
(3x + 1)(9x² - 12x + 4) = 0
(3x + 1)(3x - 2)² = 0
x = -1/3 or x = 2/3 (double roots)

That's tricky, but the rational root theorem may help

x^3 - x^2 + (4/27) = 0
27x^3 - 27x^2 + 4 = 0

Possible rational roots:

-4/27 , -2/27 , -1/27 , 1/27 , 2/27 , 4/27
-4/9 , -2/9 , -1/9 , 1/9 , 2/9 , 4/9
-4/3 , -2/3 , -1/3 , 1/3 , 2/3 , 4/3

18 possible rational roots. I'd start with the fractions that are easier to evaluate

(1/3)^3 - (1/3)^2 + 4/27
1/27 - 1/9 + 4/27
5/27 - 1/9
5/27 - 3/27
2/27

(-1/3)^3 - (-1/3)^2 + 4/27
-1/27 - 1/9 + 4/27
-4/27 + 4/27
0

x = (-1/3) is a root

x = -1/3
3x = -1
3x + 1

27x^3 - 27x^2 + 4 has a factor of (3x + 1)

(3x + 1) * (ax^2 + bx + c) = 27x^3 - 27x^2 + 0x + 4
3ax^3 + 3bx^2 + 3cx + ax^2 + bx + c = 27x^3 - 27x^2 + 0x + 4

3a = 27
3b + a = -27
3c + b = 0
c = 4

a = 9

3 * 4 + b = 0
b = -12

(3x + 1) * (9x^2 - 12x + 4) = 0

9x^2 - 12x + 4 = 0
x = (12 +/- sqrt(144 - 144)) / 18
x = (12/18)
x = 2/3

x = -1/3 , 2/3

Those are your roots.