2. For various positive numbers $a$ and $b$, it is known that

$$
a^{3}-b^{3}=3\left(2 a^{2} b-3 a b^{2}+b^{3}\right)
$$

By how many times is the larger number greater than the smaller one?
Solution. Let's consider and transform the difference:

$$
\begin{aligned}
& 0=a^{3}-b^{3}-3\left(2 a^{2} b-3 a b^{2}+b^{3}\right)= \\
& (a-b)\left(a^{2}+a b+b^{2}\right)-3\left(2 a b(a-b)-b^{2}(a-b)\right)= \\
& (a-b)\left(a^{2}+a b+b^{2}-6 a b+3 b^{2}\right)= \\
& (a-b)\left(a^{2}-5 a b+4 b^{2}\right)=(a-b)(a-4 b)(a-b)
\end{aligned}
$$

By the condition \(a \neq b\), we get \(a = 4b\), which means the larger number is 4 times greater.

Answer: 4.

Recommendations for checking. Only the correct answer - 0 points. Answer with verification - 2 points. One factor \(a-b\) is highlighted - no less than 2 points, two factors \(a-b\) are highlighted - no less than 4 points.