27 = 3 * 3 * 3

x^3 = x * x * x

1 = 1 * 1 * 1

so take the cube of both sides…

you are left with ….

3x = 1

solving for x… you get

x = 1/3

If you are looking for two imaginary roots in “a + bi” and/or “a – bi” format… the only ones that I can think of for this are…

x = 1/3 + 0i and x = 1/3 – 0i

… so that you have x = 1/3, x = 1/3 + 0i, and x = 1/3 – 0i….

…. but no matter how you write it and slice it… the answer to this problem is still… x = 1/3

A cubic function has 3 roots. Two can be imaginary, but at least one must be real.

27x^3 = 1 is the same as

27x^3 – 1 = 0

Factoring we get,

(3x – 1)^3 = 0

which is the same as

(3x – 1)(3x – 1)(3x -1) = 0

I rewrote it the second way to emphasize the fact that for this function, all 3 roots are at x = 1/3.

An example of a cubic equation with imaginary roots is

x^3 – x^2 + x -1 = 0

Factoring it we get,

(x – 1)(x^2 +1) = 0

This expression has a real root at x = 1, and two imaginary roots at x = +i and x = -i.

So going back to the original question, 27x^3 = 1 has no imaginary roots. It just has three roots, all at x = 1/3

divide both side by 27, and you get (27s will be canceled out on the left side)

x^3=1/27

then find the cube root on both side of the equation to find the value of x:

cube root of x^3 is x

cube root of 1/27 is 1/3

thus,x=1/3