Решение:
\[ (16x^3 - 32x^2) - (x - 2) = 0 \]
\[ 16x^2(x - 2) - 1(x - 2) = 0 \]
\[ (16x^2 - 1)(x - 2) = 0 \]
\[ 16x^2 - 1 = 0 \]
или
\[ x - 2 = 0 \]
\[ 16x^2 = 1 \]
\[ x^2 = \frac{1}{16} \]
\[ x = \pm\sqrt{\frac{1}{16}} = \pm\frac{1}{4} \]
\[ x = 2 \]
Ответ: $$\frac{1}{4}$$, -$$\frac{1}{4}$$, 2