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$a,$ $y^2-1+2y=2y^3$

$⇔y^2-1+2y-2y^3=0$

$⇔(y^2-1)-2y(y^2-1)=0$

$⇔(2y-1)(y^2-1)=0$

$⇔$ \(\left[ \begin{array}{l}2y-1=0\\y^2-1=0\end{array} \right.\) 

$⇔$ \(\left[ \begin{array}{l}y=\dfrac{1}{2}\\y=±1\end{array} \right.\) 

Vậy `S={\frac{1}{2};±1}`

$b,$ $(x^2+x)+(x^3+x^2)=0$

$⇔x(x+1)+x^2(x+1)=0$

$⇔(x^2+x)(x+1)=0$

$⇔x(x+1)^2=0$

$⇔$ \(\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.\) 

Vậy `S={0;-1}`.