$\frac{(x+1)^3}{3}=x^2+2x+1$
⇔ $\frac{(x+1)^3}{3}=\frac{3(x^2+2x+1)}{3}$
⇒ (x + 1)³ = 3(x² + 2x + 1)
⇔ x³ + 3x² + 3x + 1 = 3x² + 6x + 3
⇔ x³ + 3x² + 3x - 3x² - 6x + 1 - 3 = 0
⇔ x³ - 3x - 2 = 0
⇔ x³ - x - 2x - 2 = 0
⇔ x(x² - 1) - 2(x + 1) = 0
⇔ x(x - 1)(x + 1) - 2(x + 1) = 0
⇔ (x + 1)[x(x - 1) - 2] = 0
⇔ (x + 1)(x² - x - 2) = 0
⇔ (x + 1)(x² + x - 2x - 2) = 0
⇔ (x + 1)[x(x + 1) - 2(x + 1)] = 0
⇔ (x + 1)(x + 1)(x - 2) = 0
⇔ (x + 1)²(x - 2) = 0
⇔ \(\left[ \begin{array}{l}(x+1)^2=0\\x-2=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x+1=0\\x=2\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-1\\x=2\end{array} \right.\)
Vậy phương trình có S = {-1 ; 2}
