Question-157035

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

Commented by Armindo last updated on 18/Oct/21

h e l p \dots

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]{a b} + \sqrt[3]{b}}{a - b}

Commented by MJS_new last updated on 19/Oct/21

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

you think this is any better ?!

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