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Question Number 167777 by infinityaction last updated on 24/Mar/22

Answered by Jamshidbek last updated on 26/Mar/22

$$\mathbf{Solution}.$$$${\mathrm{x}}^3+{\mathrm{y}}^3+{\mathrm{z}}^3-3\mathrm{xy}=(\mathrm{x}+\mathrm{y}+\mathrm{z})({\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2-(\mathrm{xy}+\mathrm{yz}+\mathrm{xz}))$$$$\frac{1}{\sqrt[3]{\mathrm{a}}+\sqrt[3]{\mathrm{b}}+\sqrt[3]{\mathrm{c}}}=\frac{\sqrt[3]{{\mathrm{a}}^2}+\sqrt[3]{{\mathrm{b}}^2}+\sqrt[3]{{\mathrm{c}}^2}-(\sqrt[3]{\mathrm{ab}}+\sqrt[3]{\mathrm{bc}}+\sqrt[3]{\mathrm{ca}})}{\mathrm{a}+\mathrm{b}+\mathrm{c}-3\sqrt[3]{\mathrm{abc}}}=$$$$=\frac{(\sqrt[3]{{\mathrm{a}}^2}+\sqrt[3]{{\mathrm{b}}^2}+\sqrt[3]{{\mathrm{c}}^2}-(\sqrt[3]{\mathrm{ab}}+\sqrt[3]{\mathrm{bc}}+\sqrt[3]{\mathrm{ca}}))({(\mathrm{a}+\mathrm{b}+\mathrm{c})}^2+(\mathrm{a}+\mathrm{b}+\mathrm{c})\cdot 3\sqrt[3]{\mathrm{abc}}+9\sqrt[3]{{\left(\mathrm{abc}\right)}^2})}{{(\mathrm{a}+\mathrm{b}+\mathrm{c})}^3-27\mathrm{abc}}$$$$\mathrm{a}=2\mathrm{b}=3\mathrm{c}=4$$$$\mathrm{Use}\mathrm{this}\mathrm{formula}$$$$@\mathrm{math}\mathrm{\_}\mathrm{undergraduate}$$

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