a)\(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)

\(< =>\left(3x+2\right)\left(x^2-1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\)

\(< =>\left(3x+2\right)\left(x^2-1-\left(3x-2\right)\left(x+1\right)\right)=0\)

\(< =>\left(3x+2\right)\left(x^2-1-\left(3x^2+3x-2x-2\right)\right)=0\)

\(< =>\left(3x+2\right)\left(x^2-1-3x^2-3x+2x+2\right)=0\)

\(< =>\left(3x+2\right)\left(-2x^2-x+1\right)=0\)

\(< =>\left(3x+2\right)\left(-2x^2+2x-x+1\right)=0\)

\(< =>\left(3x+2\right)\left(-2x\left(x-1\right)-\left(x-1\right)\right)=0\)

\(< =>\left(3x+2\right)\left(\left(x-1\right)\left(-2x-1\right)\right)=0\)

\(< =>\left[{}\begin{matrix}3x+2=0\\x-1=0\\-2x-1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{-2}{3}\\x=1\\x=\dfrac{-1}{2}\end{matrix}\right.\)

S=\(\left\{\dfrac{-2}{3};1;\dfrac{-1}{2}\right\}\)