7. Find the real value of $x$ such that $x^{3}+3 x^{2}+3 x+7=0$.
Solution:
$$
-1-\sqrt[3]{6}
$$

Rewrite the equation as $(x+1)^{3}+6=0$ to get $(x+1)^{3}=-6 \Rightarrow x+1=\sqrt[3]{-6} \Rightarrow x=$ $-1-\sqrt[3]{6}$