Question Number 59812 by ANTARES VY last updated on 15/May/19
$$\frac{\mathrm{6}}{\mathrm{3}+\sqrt[{\mathrm{3}}]{\mathrm{3}}+\sqrt[{\mathrm{3}}]{\mathrm{9}}}\:\:\:\boldsymbol{\mathrm{simplify}}. \\ $$
Commented by Kunal12588 last updated on 15/May/19
![let (3)^(1/3) =a ⇒(9)^(1/3) =a^2 ⇒3=a^3 a+a^2 +a^3 =((a(a^3 −1))/(a−1)) [sum of GP] (6/(3+(3)^(1/3) +(9)^(1/3) )) =((2a^3 )/((a(a^3 −1))/(a−1)))=((2a^2 (a−1))/((a^3 −1)))=((2(a^3 −a^2 ))/(a^3 −1)) =((2(3−(9)^(1/3) ))/(3−1))=3−(9)^(1/3)](https://www.tinkutara.com/question/Q59817.png)
$${let}\:\sqrt[{\mathrm{3}}]{\mathrm{3}}={a} \\ $$$$\Rightarrow\sqrt[{\mathrm{3}}]{\mathrm{9}}={a}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3}={a}^{\mathrm{3}} \\ $$$${a}+{a}^{\mathrm{2}} +{a}^{\mathrm{3}} =\frac{{a}\left({a}^{\mathrm{3}} −\mathrm{1}\right)}{{a}−\mathrm{1}}\:\:\:\:\left[{sum}\:{of}\:{GP}\right] \\ $$$$\frac{\mathrm{6}}{\mathrm{3}+\sqrt[{\mathrm{3}}]{\mathrm{3}}+\sqrt[{\mathrm{3}}]{\mathrm{9}}} \\ $$$$=\frac{\mathrm{2}{a}^{\mathrm{3}} }{\frac{{a}\left({a}^{\mathrm{3}} −\mathrm{1}\right)}{{a}−\mathrm{1}}}=\frac{\mathrm{2}{a}^{\mathrm{2}} \left({a}−\mathrm{1}\right)}{\left({a}^{\mathrm{3}} −\mathrm{1}\right)}=\frac{\mathrm{2}\left({a}^{\mathrm{3}} −{a}^{\mathrm{2}} \right)}{{a}^{\mathrm{3}} −\mathrm{1}} \\ $$$$=\frac{\mathrm{2}\left(\mathrm{3}−\sqrt[{\mathrm{3}}]{\mathrm{9}}\right)}{\mathrm{3}−\mathrm{1}}=\mathrm{3}−\sqrt[{\mathrm{3}}]{\mathrm{9}} \\ $$
Answered by Kunal12588 last updated on 15/May/19
$${S}=\:\sqrt[{\mathrm{3}}]{\mathrm{3}}+\sqrt[{\mathrm{3}}]{\mathrm{9}}+\mathrm{3} \\ $$$${S}\left(\sqrt[{\mathrm{3}}]{\mathrm{3}}\right)=\sqrt[{\mathrm{3}}]{\mathrm{9}}+\mathrm{3}+\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}} \\ $$$$\Rightarrow{S}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}}}{\left(\mathrm{1}−\sqrt[{\mathrm{3}}]{\mathrm{3}}\right)}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{3}}\left(\mathrm{3}−\mathrm{1}\right)}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{1}}=\frac{\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{3}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{1}} \\ $$$$\frac{\mathrm{6}}{\mathrm{3}+\sqrt[{\mathrm{3}}]{\mathrm{3}}+\sqrt[{\mathrm{3}}]{\mathrm{9}}}=\frac{\mathrm{6}}{\frac{\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{3}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{1}}}=\frac{\mathrm{6}\left(\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{1}\right)}{\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{3}}}=\sqrt[{\mathrm{3}}]{\mathrm{9}}\left(\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{1}\right) \\ $$$$\frac{\mathrm{6}}{\mathrm{3}+\sqrt[{\mathrm{3}}]{\mathrm{3}}+\sqrt[{\mathrm{3}}]{\mathrm{9}}}=\mathrm{3}−\sqrt[{\mathrm{3}}]{\mathrm{9}} \\ $$