It looks like you've written the same polynomial equation twice. Let's focus on solving the equation:

\[ 4x^3 + 6x^2 - 12x + 8 = 0. \]

To solve this cubic equation, you can try to factor it or use techniques such as synthetic division, factoring by grouping, or the rational root theorem to find its roots.

First, let's check if there is any common factor in the polynomial:

\[ 4x^3 + 6x^2 - 12x + 8 \]

You can see that the greatest common divisor (GCD) of the coefficients (4, 6, -12, 8) is 2. Thus, factor 2 out of the equation:

\[ 2(2x^3 + 3x^2 - 6x + 4) = 0. \]

This implies that:

\[ 2x^3 + 3x^2 - 6x + 4 = 0. \]

Now, let's try using the Rational Root Theorem to find possible rational roots. The Rational Root Theorem states that any rational solution, expressed in the form of a fraction p/q, will have p as a factor of the constant term (4) and q as a factor of the leading coefficient (2). Therefore, the possible rational roots are:

\[
\pm 1, \pm 2, \pm 4, \pm \frac{1}{2}
\]

Testing these possible roots in the equation:

Let's test \( x = 1 \):

\[
2(1)^3 + 3(1)^2 - 6(1) + 4 = 2 + 3 - 6 + 4 = 3 \neq 0
\]

So, \( x = 1 \) is not a root.

Let's test \( x = -1 \):

\[
2(-1)^3 + 3(-1)^2 - 6(-1) + 4 = -2 + 3 + 6 + 4 = 11 \neq 0
\]

So, \( x = -1 \) is not a root.

Let's test \( x = 2 \):

\[
2(2)^3 + 3(2)^2 - 6(2) + 4 = 16 + 12 - 12 + 4 = 20 \neq 0
\]

So, \( x = 2 \) is not a root.

Let's test \( x = -2 \):

\[
2(-2)^3 + 3(-2)^2 - 6(-2) + 4 = -16 + 12 + 12 + 4 = 12 \neq 0
\]

So, \( x = -2 \) is not a root.

Let's test \( x = \frac{1}{2} \):

\[
2\left(\frac{1}{2}\right)^3 + 3\left(\frac{1}{2}\right)^2 - 6\left(\frac{1}{2}\right) + 4 = \frac{1}{4} + \frac{3}{4} - 3 + 4 = 2 \neq 0
\]

So, \( x = \frac{1}{2} \) is not a root.

Continue testing the remaining potential roots. Once a root is found, it can be used for synthetic division to further factor and potentially identify additional roots. If necessary, numerical or graphing methods can assist in further solving the equation after trial and error for rational roots.