Let's simplify the following fraction:

$$
\frac{x^{8}+x^{7}-x^{5}-x^{4}+x^{3}-1}{x^{6}-x^{4}+x-1}
$$
I. solution. The terms of the numerator can be paired so that the differences of the exponents in the pairs are equal (writing 1 as $x^{0}$): $8-5=7-4=3-0=3$. Accordingly - and also considering the coefficients - $x^{3}-1$ can be factored out from the numerator:

$$
\left(x^{8}-x^{5}\right)+\left(x^{7}-x^{4}\right)+\left(x^{3}-1\right)=\left(x^{3}-1\right)\left(x^{5}+x^{4}+1\right)
$$

The denominator is:

$$
x^{4}\left(x^{2}-1\right)+(x-1)=(x-1)\left[x^{4}(x+1)+1\right]=(x-1)\left(x^{5}+x^{4}+1\right),
$$

thus $x^{5}+x^{4}+1$ is a common factor. It is also known that $x^{3}-1=(x-1)\left(x^{2}+x+1\right)$, so by simplifying with $x-1$, the given expression is equal to $x^{2}+x+1$, for all $x$ for which the original denominator is not 0.

Katalin Hallósy (Szeged, Radnóti M. g. I. o. t.)

Remarks. 1. For $x=1$, $x-1=0$, and therefore the simplification makes no sense, but the expression itself does not exist either, since in this case both the numerator and the denominator are 0. Similarly, it does not exist if $x^{6}+x^{4}+1=0$. Such an $x$ exists, namely between -2 and -1, because for $x=-2$, the value of the left side is $-150$.

Mária Szaniszló (Debrecen, Mechwart A. gépip. t. I. o. t.)

2. $x^{3}-1$ can also be factored out from the denominator, indeed: $\left(x^{6}-1\right)-\left(x^{4}-x\right)=\left(x^{3}-1\right)\left(x^{3}+1-x\right)$. However, this way, we only see that the numerator is divisible by $x^{3}-x+1$ after a more complicated regrouping, because $x^{5}+x^{4}+1=\left(x^{2}+x+1\right)\left(x^{3}-x+1\right)$.

Balazs Doskar (Budapest, Piarista g. I. o. t.)

II. solution. We can try to simplify the expression by dividing the numerator by the denominator, and then only simplify the quotient of the remainder and the denominator. The division leads to a quotient of 0 and a remainder of $x^{2}+x+1$, so our expression is equal to this simpler expression.

Tibor Szilágyi (Ózd, József A. g. I. o. t.)