(3x−2)(3x+8)(x+1)2+16=0

\(\Leftrightarrow \left(\right. 9 x^{2} + 18 x - 16 \left.\right) \left(\right. x^{2} + 2 x + 1 \left.\right) + 16 = 0\)

\(\Leftrightarrow \left[\right. 9 \left(\right. x^{2} + 2 x + 1 \left.\right) - 25 \left]\right. \left(\right. x^{2} + 2 x + 1 \left.\right) + 16 = 0\)

Đặt \(x^{2} + 2 x + 1 = a \geq 0\)

\(\left(\right. 9 a - 25 \left.\right) a + 16 = 0\)

\(\Leftrightarrow 9 a^{2} - 25 a + 16 = 0\)

\(\Rightarrow \left[\right. a = 1 \\ a = \frac{16}{9}\) \(\Rightarrow \left[\right. x^{2} + 2 x + 1 = 1 \\ x^{2} + 2 x + 1 = \frac{16}{9}\) \(\Leftrightarrow \left[\right. x \left(\right. x + 2 \left.\right) = 0 \\ \left(\left(\right. x + 1 \left.\right)\right)^{2} = \left(\left(\right. \frac{4}{3} \left.\right)\right)^{2}\)

\(\Leftrightarrow \left[\right. x = 0 \\ x = - 2 \\ x + 1 = \frac{4}{3} \\ x + 1 = - \frac{4}{3}\) \(\Rightarrow \left[\right. x = 0 \\ x = - 2 \\ x = \frac{1}{3} \\ x = - \frac{7}{3}\)