8) 262 += a²b+ab² += a²b² = 5

Приведем подобные члены многочлена: $$\frac{1}{5}a^2b + \frac{2}{5}ab^2 + \frac{3}{5}ab^2 + \frac{4}{5}a^2b = (\frac{1}{5}a^2b + \frac{4}{5}a^2b) + (\frac{2}{5}ab^2 + \frac{3}{5}ab^2) = \frac{5}{5}a^2b + \frac{5}{5}ab^2 = a^2b + ab^2$$

Ответ: $$a^2b+ab^2$$