$(x-2)³+(3x+1)(3x-1)=(x+1)³$
$↔[(x-2)³-(x+1)³]+9x²-1=0$
$↔(x-2-x-1)[(x-2)²+(x-2)(x+1)+(x+1)²]+9x²-1=0$
$↔-3(x²-4x+4+x²-x-2+x²+2x+1)+9x²-1=0$
$↔-3(3x²-3x+3)+9x²-1=0$
$↔-9x²+9x-9+9x²-1=0$
$↔9x-10=0$
$↔9x=10$
$↔x=\dfrac{10}{9}$
Vậy pt có tập nghiệm $S=\bigg\{\dfrac{10}{9}\bigg\}$