25) \(\frac{5}{4}am^3-\dots\)^2 = ... - ... + \(\frac{9}{25}\)b^6...

Вспоминаем формулу квадрата разности: (a-b)² = a² - 2ab + b².
Здесь b^2 = \(\frac{9}{25}\)b^6, значит b = \(\frac{3}{5}\)b^3. И a = \(\frac{5}{4}\)am^3, тогда a² = \(\frac{5}{4}am^3\)^2=\(\frac{25}{16}\)a^2m^6 и 2ab = 2* \(\frac{5}{4}\)am^3 * \(\frac{3}{5}\)b^3 = \(\frac{3}{2}\)am^3b^3
Тогда \(\frac{5}{4}am^3 - \frac{3}{5}b^3\)^2 = \(\frac{25}{16}\)a^2m^6 - \(\frac{3}{2}\)am^3b^3 + \(\frac{9}{25}\)b^6
Ответ: \(\frac{5}{4}am^3 - \frac{3}{5}b^3\)^2 = \(\frac{25}{16}\)a^2m^6 - \(\frac{3}{2}\)am^3b^3 + \(\frac{9}{25}\)b^6