Question Number 115436 by frc2crc last updated on 25/Sep/20
$$\sqrt{\sqrt[{\mathrm{3}}]{{a}}+\sqrt[{\mathrm{3}}]{{b}}}=\frac{\mathrm{1}}{\:\sqrt{{b}−{a}×{s}^{\mathrm{3}} }}\left(\frac{−{s}^{\mathrm{2}} ×\sqrt[{\mathrm{3}}]{{a}^{\mathrm{2}} }}{\mathrm{2}}+{s}\sqrt[{\mathrm{3}}]{{ab}}+\sqrt[{\mathrm{3}}]{{b}^{\mathrm{2}} }\right) \\ $$$${what}\:{is}\:{s}? \\ $$
Answered by MJS_new last updated on 25/Sep/20
$$\mathrm{not}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{too}\:\mathrm{tired}\:\mathrm{to}\:\mathrm{type}\:\mathrm{the}\:\mathrm{path} \\ $$$${s}=−\mathrm{1}−\frac{\sqrt{{u}^{\mathrm{2}} −{uv}+{v}^{\mathrm{2}} }}{{u}}+\frac{\sqrt{\left({u}+{v}\right)\left(\mathrm{2}{u}−{v}+\mathrm{2}\sqrt{{u}^{\mathrm{2}} −{uv}+{v}^{\mathrm{2}} }\right)}}{{u}} \\ $$$$\mathrm{with}\:{u}=\sqrt[{\mathrm{3}}]{{a}}\wedge{v}=\sqrt[{\mathrm{3}}]{{b}} \\ $$$$\mathrm{true}\:\mathrm{only}\:\mathrm{for}\:{a},{b}\:\geqslant\mathrm{0}\:\mathrm{and}\:\mathrm{all}\:\mathrm{roots}\:\in\mathbb{R} \\ $$
Commented by frc2crc last updated on 25/Sep/20
$${why}\:{v}=\sqrt[{\mathrm{4}}]{{b}}? \\ $$
Commented by MJS_new last updated on 25/Sep/20
$$\mathrm{typo}\:\mathrm{sorry} \\ $$