$1.31 \frac{a^{3}-2 a^{2}+5 a+26}{a^{3}-5 a^{2}+17 a-13}$. 

Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.

$1.31 \frac{a^{3}-2 a^{2}+5 a+26}{a^{3}-5 a^{2}+17 a-13}$.
1.31 I with assistance. The numerator and denominator are transformed so that a common factor can be factored out:

$$
\begin{aligned}
& a^{3}-2 a^{2}+5 a+26=a^{3}+2 a^{2}-4 a^{2}-8 a+13 a+26= \\
& =a^{2}(a+2)-4 a(a+2)+13 a(a+2)=(a+2)\left(a^{2}-4 a+13\right) \\
& a^{3}-5 a^{2}+17 a-13=a^{3}-a^{2}-4 a^{2}+4 a+13 a-13= \\
& =(a-1)\left(a^{2}-4 a+13\right)
\end{aligned}
$$

Performing the division, we get

$$
\frac{(a+2)\left(a^{2}-4 a+13\right)}{(a-1)\left(a^{2}-4 a+13\right)}=\frac{a+2}{a-1}
$$

II with assistance. Represent the constant terms of the polynomials as $26=1 \cdot 2 \cdot 13$ and $13=1 \cdot 13$. Next, we check if each of these factors is a root of the given polynomial; for this, we substitute it with a “+” or “-” sign into the polynomial. For the polynomial $a^{3}-2 a^{2}+5 a+26$, a root is -2 (since $(-2)^{3}-2(-2)^{2}+5(-2)+26=0$). Now, by dividing the polynomial by the difference $a-(-2)$, we reduce its degree by one; we have $\left(a^{3}-2 a^{2}+5 a+26\right):(a+2)=a^{2}-4 a+13$. Thus, we obtain the factorization: $(a+2) \times \times\left(a^{2}-4 a+13\right)$. Here, the quadratic trinomial has no roots, as $D<0$. Similarly, we factorize the other polynomial and obtain the same answer.

Answer: $\frac{a+2}{a-1}$.