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Question Number 115436 by frc2crc last updated on 25/Sep/20

$$\sqrt{\sqrt[3]{a}+\sqrt[3]{b}}=\frac{1}{\sqrt{b-a\times s^3}}(\frac{-s^2\times \sqrt[3]{a^2}}{2}+s\sqrt[3]{ab}+\sqrt[3]{b^2})$$$$whatiss?$$

Answered by MJS_new last updated on 25/Sep/20

$$\mathrm{not}\mathrm{easy}\mathrm{to}\mathrm{solve}$$$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{too}\mathrm{tired}\mathrm{to}\mathrm{type}\mathrm{the}\mathrm{path}$$$$s=-1-\frac{\sqrt{u^2-uv+v^2}}{u}+\frac{\sqrt{(u+v)(2u-v+2\sqrt{u^2-uv+v^2})}}{u}$$$$\mathrm{with}u=\sqrt[3]{a}\wedge v=\sqrt[3]{b}$$$$\mathrm{true}\mathrm{only}\mathrm{for}a,b⩾0\mathrm{and}\mathrm{all}\mathrm{roots}\in \mathbb{R}$$

Commented by frc2crc last updated on 25/Sep/20

$$whyv=\sqrt[4]{b}?$$

Commented by MJS_new last updated on 25/Sep/20

$$\mathrm{typo}\mathrm{sorry}$$

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