27x^3-8=0

(difference of two cubes)

27x^3-8=0

(3x - 2)(x^2 + 6x + 4) = 0
x = 2/3 or x = ... (2 imaginary roots)

use the quadratic equation formula to find the 2 complex roots.

To solve the equation 27x^3 - 8 = 0 using the difference of two cubes factorization, follow these steps:

Step 1: Recognize the pattern:
The given equation can be seen as a difference of two cubes, where 27x^3 can be expressed as (3x)^3 and 8 as 2^3.

Step 2: Apply the difference of two cubes formula:
The difference of two cubes formula states that a^3 - b^3 = (a - b)(a^2 + ab + b^2). In this case, a = 3x and b = 2, resulting in:
(3x)^3 - 2^3 = (3x - 2)((3x)^2 + (3x)(2) + 2^2).

Step 3: Simplify:
(3x - 2)((3x)^2 + (3x)(2) + 2^2) becomes (3x - 2)(9x^2 + 6x + 4).

Step 4: Set each factor equal to zero:
We set (3x - 2) = 0 and (9x^2 + 6x + 4) = 0 to solve for x.

Step 5: Solve for x:
From (3x - 2) = 0, we can solve for x by adding 2 and dividing by 3:
3x - 2 = 0,
3x = 2,
x = 2/3.

From (9x^2 + 6x + 4) = 0, we can use the quadratic formula or factorization to solve it. However, since it does not involve a clear factorization or simple square root, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac))/2a.

For our equation, a = 9, b = 6, and c = 4. Substituting these values into the quadratic formula:

x = (-6 ± √(6^2 - 4 * 9 * 4)) / (2 * 9),
x = (-6 ± √(36 - 144)) / 18,
x = (-6 ± √(-108)) / 18.

The expression √(-108) is not a real number because the square root of a negative number is not defined in the real number system. Therefore, the equation 9x^2 + 6x + 4 = 0 does not have real solutions.

The solutions to the original equation 27x^3 - 8 = 0 are x = 2/3 (real solution) and the quadratic equation has no real solutions.