Feri wrote on the board

$$
\frac{19^{3}+65^{3}}{46^{3}+65^{3}}
$$

and "simplified" the expression by erasing the exponents, stating that the resulting expression is equal to the original one. Gábor warned him to be careful, but Feri demonstrated the equality of the two expressions with calculations and showed more similar numerical examples. Gábor is still worried that the procedure is not always valid. Can we support Feri's method? Or should we address Gábor's concerns?
Gábor's suspicion is justified, since, for example, the fraction $\left(2^{3}+1^{3}\right) /\left(3^{3}+1^{3}\right)=9 / 28$ cannot be simplified.

However, Feri was right in the given case, (1) has the value 281 484/371 961, which, when simplified by 3351, becomes 84/111, and here $84=19+65,111=46+65$.

Let's generalize the problem by replacing the numbers 19, 65, 46 with a letter each. Using known identities,

$$
\frac{a^{3}+b^{3}}{c^{3}+b^{3}}=\frac{a+b}{c+b} \cdot \frac{a^{2}-a b+b^{2}}{c^{2}-c b+b^{2}}
$$

the simplification is correct if the value of the second fraction is 1, meaning the numerator and the denominator are equal, their difference is 0, that is, if

$$
a^{2}-c^{2}-(a-c) b=(a-c)(a+c-b)=0
$$

We can assume that $a-c \neq 0$, because if $a=c$, the simplification would not be interesting. Thus, the procedure is only correct if the relationship $a+c-b=0$ holds between $a, b$, and $c$. This was true in (1), but not in the counterexample.

Michaletzky György (Budapest, Piarista g. I. o. t.)

Remark. It is not necessary to separately state that the simplifying factor must be different from 0, because

$$
c^{2}-c b+b^{2}=\left(c-\frac{b}{2}\right)^{2}+\frac{3 b^{2}}{4}
$$

which only becomes 0 if $c=b=0$.