Решение:

a) $$a^{-6} \cdot a^{7} = a^{-6+7} = a^{1} = a$$

б) $$a^{-8} : a^{3} = a^{-8-3} = a^{-11} = \frac{1}{a^{11}}$$

в) $$(a^{3})^{-6} \cdot a^{21} = a^{3 \cdot (-6)} \cdot a^{21} = a^{-18} \cdot a^{21} = a^{-18+21} = a^{3}$$

г) $$3\sqrt{24} + \sqrt{54} = 3\sqrt{4 \cdot 6} + \sqrt{9 \cdot 6} = 3 \cdot 2\sqrt{6} + 3\sqrt{6} = 6\sqrt{6} + 3\sqrt{6} = 9\sqrt{6}$$

д) $$\frac{5\sqrt{3} \cdot \sqrt{15}}{\sqrt{5}} = \frac{5\sqrt{3 \cdot 15}}{\sqrt{5}} = \frac{5\sqrt{45}}{\sqrt{5}} = 5\sqrt{\frac{45}{5}} = 5\sqrt{9} = 5 \cdot 3 = 15$$

e) $$(2\sqrt{5} + 3)^{2} = (2\sqrt{5})^{2} + 2 \cdot 2\sqrt{5} \cdot 3 + 3^{2} = 4 \cdot 5 + 12\sqrt{5} + 9 = 20 + 12\sqrt{5} + 9 = 29 + 12\sqrt{5}$$