To prove that since $a b(a+b)$ is divisible by $a^{2}+a b+b^{2}$, we first divide $a b(a+b)$ by $a^{2}+a b+b^{2}$, obtaining
$$a b(a+b)=\left(a^{2}+a b+b^{2}\right) a-a^{3}$$

Thus, $\left(a^{2}+a b+b^{2}\right) \mid a^{3}$. Similarly, $\left(a^{2}+a b+b^{2}\right) \mid b^{3}$, meaning $a^{2}+a b+b^{2}$ is a common divisor of $a^{3}$ and $b^{3}$, so $\left(a^{2}+a b+b^{2}\right) \mid\left(a^{3}, b^{3}\right)$. (See (3) in Unit 2.) Also, $\left(a^{3}, b^{3}\right)=(a, b)^{3}$ (see the note below), hence
$$\left(a^{2}+a b+b^{2}\right) \mid(a, b)^{3} .$$

Let $d=(a, b), a=a_{1} d, b=b_{1} d$, then (1) becomes $\left(a_{1}^{2}+a_{1} b_{1}+b_{1}^{2}\right) \mid d$. Therefore, $d \geqslant a_{1}^{2}+a_{1} b_{1}+b_{1}^{2}$, and more specifically, $d > a_{1} b_{1}$. Since $a \neq b$, the integers $a_{1} \neq b_{1}$, thus $\left|a_{1}-b_{1}\right| \geqslant 1$, leading to
$$|a-b|^{3}=d^{3}\left|a_{1}-b_{1}\right|^{3} \geqslant d^{3}>d^{2} a_{1} b_{1}=a b$$

Thus, $|a-b|>\sqrt[3]{a b}$.