To solve the equation 8x^3 - 12x^2 = -6x + 1, we can start by rearranging the terms to get all the terms on one side of the equation:

8x^3 - 12x^2 + 6x - 1 = 0

Now, we can try to factor the equation. We notice that the coefficients of the terms are all multiples of 2, so we can factor out a 2:

2(4x^3 - 6x^2 + 3x - 1) = 0

Now, we can try to factor the expression inside the parentheses. Since the degree of the polynomial is 3, we can use the Rational Root Theorem to find possible rational roots. The possible rational roots are the factors of the constant term (in this case, 1) divided by the factors of the leading coefficient (in this case, 4).

The factors of 1 are ±1, and the factors of 4 are ±1, ±2, and ±4. So, the possible rational roots are ±1/4, ±1/2, ±1, ±2, ±1/4, ±1/2, ±1, ±2.

We can try these values one by one to see if any of them are roots of the polynomial. By trying these values, we find that x = 1/2 is a root of the polynomial.

Therefore, we can divide the polynomial by (x - 1/2) using synthetic division:

1/2 | 4 -6 3 -1
2 -2/3 7/6
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4 -4 7/3 1/6

The result is 4x^2 - 4x + 7/3 + 1/6.

Now, we have a quadratic equation. We can solve this equation by factoring or using the quadratic formula. However, the quadratic equation does not factor nicely, so we will use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 4, b = -4, and c = 7/3 + 1/6.

x = (-(-4) ± √((-4)^2 - 4(4)(7/3 + 1/6))) / (2(4))
= (4 ± √(16 - 4(4)(7/3 + 1/6))) / 8
= (4 ± √(16 - 4(4)(21/6 + 1/6))) / 8
= (4 ± √(16 - 4(4)(22/6))) / 8
= (4 ± √(16 - 4(88/6))) / 8
= (4 ± √(16 - 704/6)) / 8
= (4 ± √(16 - 352/3)) / 8
= (4 ± √(48/3 - 352/3)) / 8
= (4 ± √(-304/3)) / 8

Since the discriminant (b^2 - 4ac) is negative, the quadratic equation has no real solutions. Therefore, the equation 8x^3 - 12x^2 = -6x + 1 has only one real root, which is x = 1/2.