[tex] \frac{6}{ \sqrt[3]{5} +1} [/tex] [tex] \frac{3}{ \sqrt[3]{16}+ \sqrt[3]{4}+ 1} [/tex]. Created by Lenalook14. algebra-ru

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[tex] \frac{6}{ \sqrt[3]{5} +1} [/tex]
[tex] \frac{3}{ \sqrt[3]{16}+ \sqrt[3]{4}+ 1} [/tex]

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1) $\frac{6*( \sqrt[3]{25}- \sqrt[3]{5} +1) }{( \sqrt[3]{5} +1)( \sqrt[3]{25}- \sqrt[3]{5} +1) }= \frac{6*( \sqrt[3]{25}- \sqrt[3]{5}+1) }{( \sqrt[3]{5}) ^{3} +1 ^{3} } = \frac{6*( \sqrt[3]{25} - \sqrt[3]{5} +1)}{5+1} = \frac{6( \sqrt[3]{25}- \sqrt[3]{5}+1) }{6} = \sqrt[3]{25}- \sqrt[3]{5} +1$
2) $\frac{3( \sqrt[3]{4}-1)}{( \sqrt[3]{4}-1)( \sqrt[3]{16}+ \sqrt[3]{4}+1) }= \frac{3( \sqrt[3]{4}-1) }{( \sqrt[3]{4} ) ^{3}- 1 ^{3} } = \frac{3( \sqrt[3]{4} -1)}{4-1}= \frac{3( \sqrt[3]{4}-1) }{3} = \sqrt[3]{4}-1$

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