Solve the equation $(2x - 3b)^2 = (x + 6b)(x - 9b)(8x - 12b) - (3b)^3$ for $x$.

1. Problem Description

Solve the equation $(2x - 3b)^2 = (x + 6b)(x - 9b)(8x - 12b) - (3b)^3$ for $x$.

2. Solution Steps

First, expand the left side of the equation:

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

Next, expand the right side of the equation:

$(x + 6b)(x - 9b) = x^2 - 9bx + 6bx - 54b^2 = x^2 - 3bx - 54b^2$

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

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

So the right side of the equation becomes:

$8x^3 - 36bx^2 - 396b^2x + 648b^3 - 27b^3 = 8x^3 - 36bx^2 - 396b^2x + 621b^3$

Now, the equation is:

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

Rearrange the equation:

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

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

However, note that $8x - 12b = 4(2x - 3b)$. Thus, the equation becomes:

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

$(2x - 3b)^2 - (x + 6b)(x - 9b)4(2x - 3b) + (3b)^3 = 0$

Let $2x - 3b = y$. Then $2x = y + 3b$, $x = \frac{y}{2} + \frac{3b}{2}$

$(\frac{y}{2} + \frac{3b}{2} + 6b)(\frac{y}{2} + \frac{3b}{2} - 9b) = (\frac{y}{2} + \frac{15b}{2})(\frac{y}{2} - \frac{15b}{2}) = \frac{y^2}{4} - \frac{225b^2}{4}$

Thus, the equation becomes:

$y^2 - 4(\frac{y^2}{4} - \frac{225b^2}{4})y + 27b^3 = 0$

$y^2 - (y^2 - 225b^2)4y + 27b^3 = 0$

$y^2 - (4y^3 - 900b^2 y) + 27b^3 = 0$

$-4y^3 + y^2 + 900b^2 y + 27b^3 = 0$

$y=9b$ is a solution because $-4(9b)^3 + (9b)^2 + 900b^2(9b) + 27b^3 = -2916b^3 + 81b^2 + 8100b^3 + 27b^3 = 5211b^3+81b^2 \ne 0$

Let's rearrange the initial equation:

$4x^2 - 12bx + 9b^2 = (x + 6b)(x - 9b)(8x - 12b) - 27b^3$

$4x^2 - 12bx + 9b^2 = 4(x + 6b)(x - 9b)(2x - 3b) - 27b^3$

If $x = \frac{3}{2}b$ then $4(\frac{9}{4}b^2) - 12b(\frac{3}{2}b) + 9b^2 = 9b^2 - 18b^2 + 9b^2 = 0$

and $(x+6b)(x-9b)(8x-12b) - 27b^3 = (3/2 b + 6b)(3/2b - 9b)(12b-12b)-27b^3 = (15/2 b)(-15/2b)(0) -27b^3 = -27b^3$

Thus, $x=\frac{3}{2}b$ implies that $0=-27b^3$, so $b=0$, in which case $x$ can be any value.

Let $2x - 3b = 0$, so $x = \frac{3}{2}b$. If $x = 6b$, $(x + 6b)(x - 9b)(8x - 12b) - (3b)^3 = (12b)(-3b)(36b) - 27b^3 = -1296b^3 - 27b^3 = -1323b^3$

$(2x - 3b)^2 = (12b - 3b)^2 = (9b)^2 = 81b^2$, so $81b^2 = -1323b^3$, implies $81 = -1323b$, so $b = -81/1323 = -3/49$

Then $x = 6b = -18/49$.

Consider $(2x - 3b)^2 = 4x^2 - 12bx + 9b^2$.

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

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

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

If $x = 9b$, $8(9b)^3 - 36b(9b)^2 - 4(9b)^2 - 396b^2(9b) + 12b(9b) + 621b^3 - 9b^2 = 5832b^3 - 2916b^3 - 324b^2 - 3564b^3 + 108b^2 + 621b^3 - 9b^2 = -1b^2$

Rewrite $(2x - 3b)^2 = (x+6b)(x-9b)(8x-12b)-(3b)^3$ as $(2x-3b)^2 = 4(x+6b)(x-9b)(2x-3b)-(3b)^3$

If $2x-3b = 0$, then $0 = -27b^3$. If $b\ne 0$ then $2x-3b\ne 0$.

Consider $x = 9/2 b$

LHS: $(9b - 3b)^2 = 36b^2$.

RHS: $(9/2b+6b)(9/2b-9b)(36b-12b)-27b^3 = (21/2b)(-9/2b)(24b)-27b^3 = \frac{-4536}{4}b^3 - 27b^3 = -1134 b^3 - 27 b^3 = -1161b^3$

$36b^2 = -1161b^3$, therefore $36=-1161b$

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

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

If $x = 3b$: $8*27b^3 - 9b^2(36b + 4) + 3b(12b + 396b^2) + 621b^3 - 9b^2 = 216b^3 - 324b^3 - 36b^2 + 36b^2 + 1188b^3 + 621b^3 - 9b^2= 1641b^3 - 9b^2$

3. Final Answer

The solution is complex and difficult to express in a closed form.

The equation can be simplified to $8x^3-(36b+4)x^2+(12b-396b^2)x+(621b^3-9b^2)=0$. There is no easy solution to obtain for this cubic equation. If $b=0$, then $8x^3-4x^2 = 0$ thus $4x^2(2x-1) = 0$, meaning $x=0$ or $x=1/2$.