a) \(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\left(x+2\right)\)
\(\Leftrightarrow\left(x^3-3x^2+3x-1\right)+3\left(x^2+2x+1\right)=x^3+8\)
\(\Leftrightarrow x^3-3x^2+3x-1+3x^2+6x+3=x^3+8\)
\(\Leftrightarrow x^3-3x^2+3x+3x^2+6x-x^3=1-3+8\)
\(\Leftrightarrow9x=6\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
Vậy \(S=\left\{\dfrac{2}{3}\right\}\)
b) \(3\left(x-2\right)^2+\left(x-1\right)^3-x^3=7\)
\(\Leftrightarrow3\left(x^2-4x+4\right)+\left(x^2-3x^2+3x-1\right)-x^3=7\)
\(\Leftrightarrow3x^2-12x+12+x^3-3x^2+3x-1-x^3=7\)
\(\Leftrightarrow12x=-4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Vậy \(S=\left\{-\dfrac{1}{3}\right\}\)