OpenSAT

The expression $x^3 + 3x^2 + 3x + 1$ is a perfect cube: $(x+1)^3$. Since $x^2 + 2x + 1 = 0$ is equivalent to $(x + 1)^2 = 0$, it follows that $x + 1 = 0$, and therefore, $(x + 1)^3 = 0^3 = 0$. Thus, the value of $x^3 + 3x^2 + 3x + 1$ is 0.