The problem asks us to solve for $x$ in the following equation: $(2x - 3b)^2 = (x + 6b)(x - 9b)(8x - 12b) - (3b)^3$

1. Problem Description

The problem asks us to solve for $x$ in the following equation:

$(2x - 3b)^2 = (x + 6b)(x - 9b)(8x - 12b) - (3b)^3$

2. Solution Steps

First, expand both sides of the equation:

$(2x - 3b)^2 = 4x^2 - 12xb + 9b^2$

$(x + 6b)(x - 9b)(8x - 12b) = (x^2 - 9xb + 6xb - 54b^2)(8x - 12b) = (x^2 - 3xb - 54b^2)(8x - 12b) = 8x^3 - 12x^2b - 24x^2b + 36xb^2 - 432xb^2 + 648b^3 = 8x^3 - 36x^2b - 396xb^2 + 648b^3$

$(3b)^3 = 27b^3$

Now we can substitute these expansions into the original equation:

$4x^2 - 12xb + 9b^2 = 8x^3 - 36x^2b - 396xb^2 + 648b^3 - 27b^3$

$4x^2 - 12xb + 9b^2 = 8x^3 - 36x^2b - 396xb^2 + 621b^3$

Rearrange the equation to set it to zero:

$0 = 8x^3 - 36x^2b - 4x^2 - 396xb^2 + 12xb + 621b^3 - 9b^2$

$0 = 8x^3 - (36b + 4)x^2 + (12b - 396b^2)x + (621b^3 - 9b^2)$

This equation looks complicated. Let's reconsider our previous steps. The term $(x+6b)(x-9b)(8x-12b) = (x+6b)(x-9b)4(2x-3b)$. Notice that $(2x-3b)$ is present in the expansion of the left-hand side as well.

Let's write $(x + 6b)(x - 9b)(8x - 12b) = (x + 6b)(x - 9b)4(2x - 3b) = 4(x^2 - 3xb - 54b^2)(2x - 3b) = 4(2x^3 - 3x^2b - 6x^2b + 9xb^2 - 108xb^2 + 162b^3) = 4(2x^3 - 9x^2b - 99xb^2 + 162b^3) = 8x^3 - 36x^2b - 396xb^2 + 648b^3$

So, we have $4x^2 - 12xb + 9b^2 = 8x^3 - 36x^2b - 396xb^2 + 648b^3 - 27b^3$

$4x^2 - 12xb + 9b^2 = 8x^3 - 36x^2b - 396xb^2 + 621b^3$

Rearranging:

$8x^3 - 36x^2b - 4x^2 - 396xb^2 + 12xb + 621b^3 - 9b^2 = 0$

$8x^3 - (36b+4)x^2 + (12b-396b^2)x + (621b^3-9b^2) = 0$

If the equation was $(2x-3b)^2 = (x+6b)(x-9b)(8x-12b)-(3b)^3$, let us rewrite the right side:

$4(x+6b)(x-9b)(2x-3b) - 27b^3$.

Let $2x - 3b = y$. Then $x = (y+3b)/2$.

$(x+6b) = (y+3b)/2 + 6b = (y+15b)/2$

$(x-9b) = (y+3b)/2 - 9b = (y-15b)/2$

$y^2 = 4 ( \frac{y+15b}{2} ) ( \frac{y-15b}{2} ) y - 27b^3$

$y^2 = (y^2 - 225b^2)y - 27b^3$

$y^2 = y^3 - 225b^2 y - 27b^3$

$y^3 - y^2 - 225b^2 y - 27b^3 = 0$

We can express everything in terms of $y$ now. Let us go back to the original equation.

$(2x - 3b)^2 = (x + 6b)(x - 9b)(8x - 12b) - (3b)^3$

$(2x - 3b)^2 = (x + 6b)(x - 9b)4(2x - 3b) - 27b^3$

Let $y = 2x - 3b$. Thus, $x = (y+3b)/2$

Then $(x+6b) = (y+3b)/2 + 6b = (y + 15b)/2$

And $(x - 9b) = (y+3b)/2 - 9b = (y - 15b)/2$

$y^2 = (\frac{y+15b}{2})(\frac{y-15b}{2})4y - 27b^3$

$y^2 = (y^2 - 225b^2)y - 27b^3$

$y^2 = y^3 - 225b^2y - 27b^3$

$y^3 - y^2 - 225b^2 y - 27b^3 = 0$.

If we consider $y = 15b$, $15^{3}b^3 - 15^{2}b^2 - 225b^{2}(15b) - 27b^{3} = 3375b^{3} - 225b^{2} - 3375b^{3} - 27b^{3} \neq 0$

If we consider $y = -3b$, $(-3b)^3 - (-3b)^2 - 225b^2(-3b) - 27b^3 = -27b^3 - 9b^2 + 675b^3 - 27b^3 = 621b^3 - 9b^2 \neq 0$

Let's try $2x - 3b = 0$, $x = 3b/2$.

$(0) = (\frac{3b}{2} + 6b)(\frac{3b}{2} - 9b)4(\frac{3b}{2} - \frac{3b}{2}) - 27b^3$.

$0 = (\frac{15b}{2})(\frac{-15b}{2})(0) - 27b^3$

$0 = 0 - 27b^3$

$-27b^3 = 0$, so $b = 0$.

When $b = 0$, $(2x)^2 = x*x*8x - 0$

$4x^2 = 8x^3$, $4x^2(2x - 1) = 0$

$x = 0$ or $x = 1/2$.

If b = 0, the equation reduces to $4x^2 = 8x^3$, which simplifies to $2x^3 - x^2 = 0$, so $x^2(2x-1)=0$. Thus $x=0$ or $x=1/2$.

3. Final Answer

If $b = 0$, then $x = 0$ or $x = 1/2$. Otherwise, it's too complicated to solve.

If b=0: x = 0, x = 1/2