Let's simplify the following fraction:

$$
\frac{\left(a^{3}+b^{3}\right)(a+b)^{3}+2 a^{3} b^{3}}{\left(a^{2}+b^{2}\right)(a+b)^{2}+a^{2} b^{3}}
$$
The two factors in the denominator are non-negative, and only become 0 when $a=b=0$. This is the only pair of values for $a, b$ for which the expression is undefined, and we exclude it.

The denominator in the following form:

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

by factoring out the common factor of the 2nd and 3rd terms in the parentheses, the 2nd factor of the 1st term in the parentheses is obtained, thus

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

Here, the expression in the parentheses cannot be further factored, because we have seen that it is only 0 when $a=b=0$, and if it could be written as the product of two (linear) factors, then there would be a pair of values for $a, b$ for which the value of one of the factors would be 0, and thus the product would also vanish.

Accordingly, the fraction can only be simplified by $M=a^{2}+a b+b^{2}$ or by $N=M^{2}$ itself. Testing with the easily calculable pair of values $a=2, b=1$, we get $M=7$, and the numerator is $9 \cdot 27+2 \cdot 8=259=7 \cdot 37$, which suggests that we can simplify by $M$. However, it is not expected that we can simplify by $M^{2}$, since $7 \cdot 37$ is not divisible by $7^{2}$.

The expansion of the numerator and appropriate grouping confirms our conjecture:

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

However, from the new parentheses, $M$ cannot be factored out, because it can be written as:

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

and here the 2nd term is not divisible by $M$. Therefore, the simplified form of the fraction is:

$$
\frac{a^{4}+2 a^{3} b+2 a b^{3}+b^{4}}{a^{2}+a b+b^{2}}
$$

János Szeredi (Budapest, II. Rákóczi F. Gymnasium, 2nd year)

György Kiss (Győr, Czuczor G. Bencés Gymnasium, 2nd year)

Remark. The factorizations can easily be derived from the observation that the value of the expression does not change if $a$ and $b$ are swapped. In such cases, it is useful to take the sum $a+b=s$ and the product $a b=p$ of the original variables as new variables. With these, the denominator is

$$
N=\left(s^{2}-2 p\right) s^{2}+p^{2}=s^{4}-2 p s^{2}+p^{2}=\left(s^{2}-p\right)^{2}=\left(a^{2}+a b+b^{2}\right)^{2}
$$

and the numerator is

$$
\begin{aligned}
\left(s^{2}-3 p\right) s^{4}+2 p^{3} & =s^{6}-3 s^{4} p+2 p^{3}=s^{4}\left(s^{2}-p\right)-2 p\left(s^{4}-p^{2}\right)= \\
& =\left(s^{2}-p\right)\left\{s^{4}-2 p\left(s^{2}+p\right)\right\}
\end{aligned}
$$