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Question Number 170945 by Rasheed.Sindhi last updated on 04/Jun/22

y=(9)^(1/3) +(3)^(1/3) +4 ; (1+(1/y))^3 =?

$$ \\ $$$${y}=\sqrt[{\mathrm{3}}]{\mathrm{9}}\:+\sqrt[{\mathrm{3}}]{\mathrm{3}}\:+\mathrm{4}\:\:\:;\:\:\:\left(\mathrm{1}+\frac{\mathrm{1}}{{y}}\right)^{\mathrm{3}} =? \\ $$$$ \\ $$

Answered by cortano1 last updated on 04/Jun/22

y−3=(9)^(1/3) +(3)^(1/3) +1 (y−3)((3)^(1/3) −1)=2 y=(2/( (3)^(1/3) −1))+3=((3(3)^(1/3) −1)/( (3)^(1/3) −1)) (1+(1/y))^3 =(1+(((3)^(1/3) −1)/(3(3)^(1/3) −1)))^3 =(((4(3)^(1/3) −2)/(3(3)^(1/3) −1)))^3 =((48(3)^(1/3) −96(9)^(1/3) +184)/(9(3)^(1/3) −27(9)^(1/3) +80))

$$\:{y}−\mathrm{3}=\sqrt[{\mathrm{3}}]{\mathrm{9}}\:+\sqrt[{\mathrm{3}}]{\mathrm{3}}\:+\mathrm{1} \\ $$$$\left({y}−\mathrm{3}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{1}\right)=\mathrm{2} \\ $$$${y}=\frac{\mathrm{2}}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{1}}+\mathrm{3}=\frac{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{1}} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{y}}\right)^{\mathrm{3}} =\left(\mathrm{1}+\frac{\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{1}}{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{1}}\right)^{\mathrm{3}} \\ $$$$=\left(\frac{\mathrm{4}\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{2}}{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{1}}\right)^{\mathrm{3}} \:=\frac{\mathrm{48}\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{96}\sqrt[{\mathrm{3}}]{\mathrm{9}}+\mathrm{184}}{\mathrm{9}\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{27}\sqrt[{\mathrm{3}}]{\mathrm{9}}+\mathrm{80}} \\ $$

Commented by Rasheed.Sindhi last updated on 06/Jun/22

Thank you sir!

$$\mathcal{T}{hank}\:{you}\:{sir}! \\ $$

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