Помогите решить 8, 9, 10

Решите задачу:

$8) (5+17 ^{ \frac{1}{2} } ) ^{- \frac{1}{3} } :(5- \sqrt{17} ) ^{ \frac{1}{3} } =(5+ \sqrt{17} ) ^{- \frac{1}{3} } :(5- \sqrt{17} ) ^{ \frac{1}{3} } = \frac{1}{(5+ \sqrt{17} ) ^{ \frac{1}{3} } } : \\ \\ :(5- \sqrt{17} ) ^{ \frac{1}{3} } = \frac{1}{(5+ \sqrt{17} ) ^{ \frac{1}{3} } } * \frac{1}{(5- \sqrt{17} ) ^{ \frac{1}{3} } } = \frac{1}{(25-17) ^{ \frac{1}{3} } } = \frac{1}{ \sqrt[3]{8} } = \frac{1}{2}$

$9) \sqrt{ \frac{ \sqrt{56 ^{2} -46 ^{2} } }{0.25* \sqrt{10} } } = \sqrt{ \frac{ \sqrt{(56-46)(56+46)} }{0.25* \sqrt{10} } } = \sqrt{ \frac{ \sqrt{10*102} }{0.25* \sqrt{10} } } = \sqrt{ \frac{ \sqrt{10} * \sqrt{102} }{0.25* \sqrt{10} } } = \\ \\ = \sqrt{ \frac{ \sqrt{102} }{0.25} } = \sqrt{ \frac{ \sqrt{102} }{ \frac{1}{4} } } = \sqrt{4 \sqrt{102} } =2 \sqrt[4]{102}$

$10) \frac{7( \sqrt{5} - \sqrt{3} ) ^{2} }{20(2+ \sqrt[4]{15})(2- \sqrt[4]{15}) } = \frac{7(5+3-2 \sqrt{15}) }{20(4- \sqrt[2]{15}) } = \frac{7(8-2 \sqrt{15}) }{20(4- \sqrt{15} )} = \frac{7*2(4- \sqrt{15}) }{20(4- \sqrt{15}) } = \frac{7*2}{20} = \\\\= \frac{7}{10}=0.7$