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Question Number 157035 by Armindo last updated on 18/Oct/21

Commented by Armindo last updated on 18/Oct/21

$$help...$$

Commented by MJS_new last updated on 18/Oct/21

$$\frac{1}{a^{1/3}+b^{1/3}}=\frac{a^{2/3}-a^{1/3}{b}^{1/3}+b^{2/3}}{a+b}$$

Commented by cortano last updated on 19/Oct/21

$$\frac{1}{\sqrt[3]{a}-\sqrt[3]{b}}=\frac{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b}}{a-b}$$

Commented by MJS_new last updated on 19/Oct/21

$$\mathrm{yes}.\mathrm{for}\mathrm{real}\mathrm{numbers}\sqrt[3]{-x}=-\sqrt[3]{x}\mathrm{so}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{same}$$

Answered by depressiveshrek last updated on 18/Oct/21

$$\frac{1}{\sqrt[3]{15}-\sqrt[3]{7}}$$$$$$$$\frac{1}{\sqrt[3]{15}-\sqrt[3]{7}}\times \frac{\sqrt[3]{15}+\sqrt[3]{7}}{\sqrt[1]{15}+\sqrt[3]{7}}$$$$$$$$\frac{\sqrt[3]{15}+\sqrt[3]{7}}{(\sqrt[3]{15}-\sqrt[3]{7})(\sqrt[3]{15}+\sqrt[3]{7})}$$$$$$$$\frac{\sqrt[3]{15}+\sqrt[3]{7}}{\sqrt[3]{{15}^2}-\sqrt[3]{7^2}}$$$$$$$$\frac{\sqrt[3]{15}+\sqrt[3]{7}}{\sqrt[3]{225}-\sqrt[3]{49}}$$$$$$

Commented by MJS_new last updated on 18/Oct/21

$$\mathrm{you}\mathrm{think}\mathrm{this}\mathrm{is}\mathrm{any}\mathrm{better}?!$$

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