8.4. Find the natural number $x$ that satisfies the equation

$$
x^{3}=2011^{2}+2011 \cdot 2012+2012^{2}+2011^{3}
$$
Answer: 2012. Solution: Let $a=2011, \quad b=2012$. Then $a^{2}+a b+b^{2}=\frac{b^{3}-a^{3}}{b-a}=b^{3}-a^{3}$ (since $b-a=1$). Therefore, $x^{3}=b^{3}-a^{3}+a^{3}=b^{3}$. Hence, $x=b=2012$.