$(3x - 2)³ - (x - 3)³ = (2x + 1)³$
$⇔ (2x + 1)³ + (x - 3)³ = (3x - 2)³$
$⇔ [(2x + 1) + (x - 3)][(2x + 1)² - (2x + 1)(x - 3) + (x - 3)²] - (3x - 2)³ = 0$
$⇔ (3x - 2)(4x² + 4x + 1 - 2x² + 6x - x + 3 + x² - 6x + 9) - (3x - 2)(9x² - 12x + 4) = 0$
$⇔ (3x - 2)(3x² + 3x + 13 - 9x² + 12x - 4) = 0$
$⇔ (3x - 2)(-6x² + 15x + 9) = 0$
$⇔ (3x - 2)(x - 3)(2x + 1) = 0$
$⇔\left[ \begin{array}{l}3x-2=0\\x-3=0\\2x+1=0\end{array} \right.$
$⇔\left[ \begin{array}{l}x=\frac{2}{3}\\x=3\\2x=-\frac{1}{2}\end{array} \right.$
