Разберем каждую часть отдельно: 1)а⁵b³-4a³b=a³b(a²b²-4)=a³b((ab)²-4)=a³b(ab-2)(ab+2) 2)a²b-2a³=a²(b-2a) Ceйчас разделим:
![\frac{( \sqrt[3]{64})^3-4* \sqrt[3]{64})}{ \sqrt[3]{2}-2 \sqrt[3]{32}}= \frac{64-16}{ \sqrt[3]{2}- \sqrt[3]{256}}= \frac{48}{ \sqrt[3]{2}(1- \sqrt[3]{128})}](https://tex.z-dn.net/?f=+%5Cfrac%7B%28+%5Csqrt%5B3%5D%7B64%7D%29%5E3-4%2A+%5Csqrt%5B3%5D%7B64%7D%29%7D%7B+%5Csqrt%5B3%5D%7B2%7D-2+%5Csqrt%5B3%5D%7B32%7D%7D%3D+%5Cfrac%7B64-16%7D%7B+%5Csqrt%5B3%5D%7B2%7D-++%5Csqrt%5B3%5D%7B256%7D%7D%3D+%5Cfrac%7B48%7D%7B+%5Csqrt%5B3%5D%7B2%7D%281-+%5Csqrt%5B3%5D%7B128%7D%29%7D "\frac{( \sqrt[3]{64})^3-4* \sqrt[3]{64})}{ \sqrt[3]{2}-2 \sqrt[3]{32}}= \frac{64-16}{ \sqrt[3]{2}- \sqrt[3]{256}}= \frac{48}{ \sqrt[3]{2}(1- \sqrt[3]{128})}")