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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...

$${help}... \\ $$

Commented by MJS_new last updated on 18/Oct/21

(1/(a^(1/3) +b^(1/3) ))=((a^(2/3) −a^(1/3) b^(1/3) +b^(2/3) )/(a+b))

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

Commented by cortano last updated on 19/Oct/21

 (1/( (a)^(1/3) −(b)^(1/3) )) = (((a^2 )^(1/3) +((ab))^(1/3) +(b)^(1/3) )/(a−b))

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

Commented by MJS_new last updated on 19/Oct/21

yes. for real numbers ((−x))^(1/3) =−(x)^(1/3)  so it′s the same

$$\mathrm{yes}.\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}\:\sqrt[{\mathrm{3}}]{−{x}}=−\sqrt[{\mathrm{3}}]{{x}}\:\mathrm{so}\:\mathrm{it}'\mathrm{s}\:\mathrm{the}\:\mathrm{same} \\ $$

Answered by depressiveshrek last updated on 18/Oct/21

(1/( ((15))^(1/3) −(7)^(1/3) ))     (1/( ((15))^(1/3) −(7)^(1/3) ))×((((15))^(1/3) +(7)^(1/3) )/( ((15))^(1/1) +(7)^(1/3) ))     ((((15))^(1/3) +(7)^(1/3) )/( (((15))^(1/3) −(7)^(1/3) )(((15))^(1/3) +(7)^(1/3) )))     ((((15))^(1/3) +(7)^(1/3) )/( ((15^2 ))^(1/3) −(7^2 )^(1/3) ))     ((((15))^(1/3) +(7)^(1/3) )/( ((225))^(1/3) −((49))^(1/3) ))

$$\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{15}}−\sqrt[{\mathrm{3}}]{\mathrm{7}}} \\ $$$$\: \\ $$$$\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{15}}−\sqrt[{\mathrm{3}}]{\mathrm{7}}}×\frac{\sqrt[{\mathrm{3}}]{\mathrm{15}}+\sqrt[{\mathrm{3}}]{\mathrm{7}}}{\:\sqrt[{\mathrm{1}}]{\mathrm{15}}+\sqrt[{\mathrm{3}}]{\mathrm{7}}} \\ $$$$\: \\ $$$$\frac{\sqrt[{\mathrm{3}}]{\mathrm{15}}+\sqrt[{\mathrm{3}}]{\mathrm{7}}}{\:\left(\sqrt[{\mathrm{3}}]{\mathrm{15}}−\sqrt[{\mathrm{3}}]{\mathrm{7}}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{15}}+\sqrt[{\mathrm{3}}]{\mathrm{7}}\right)} \\ $$$$\: \\ $$$$\frac{\sqrt[{\mathrm{3}}]{\mathrm{15}}+\sqrt[{\mathrm{3}}]{\mathrm{7}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{15}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{\mathrm{7}^{\mathrm{2}} }} \\ $$$$\: \\ $$$$\frac{\sqrt[{\mathrm{3}}]{\mathrm{15}}+\sqrt[{\mathrm{3}}]{\mathrm{7}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{225}}−\sqrt[{\mathrm{3}}]{\mathrm{49}}} \\ $$$$\: \\ $$

Commented by MJS_new last updated on 18/Oct/21

you think this is any better?!

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

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