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Question Number 94257 by pete last updated on 17/May/20

$$\mathrm{If}{9\mathrm{y}}^2+\frac{1}{{\mathrm{y}}^2}=3,\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}$$$${27\mathrm{y}}^3+\frac{1}{{\mathrm{y}}^3}$$

Answered by niroj last updated on 17/May/20

$${(3\mathrm{y}+\frac{1}{\mathrm{y}})}^2-2.3\mathrm{y}.\frac{1}{\mathrm{y}}={9\mathrm{y}}^2+\frac{1}{{\mathrm{y}}^2}$$$${(3\mathrm{y}+\frac{1}{\mathrm{y}})}^2=3+6$$$$3\mathrm{y}+\frac{1}{\mathrm{y}}=3$$$$\mathrm{now},$$$${(3\mathrm{y}+\frac{1}{\mathrm{y}})}^3={\left(3\mathrm{y}\right)}^3+{\left(\frac{1}{\mathrm{y}}\right)}^3+3.3\mathrm{y}.\frac{1}{\mathrm{y}}(3\mathrm{y}+\frac{1}{\mathrm{y}})$$$$={27\mathrm{y}}^3+\frac{1}{{\mathrm{y}}^3}+9.3$$$${27\mathrm{y}}^3+\frac{1}{{\mathrm{y}}^3}={(3\mathrm{y}+\frac{1}{\mathrm{y}})}^3-27$$$$=3^3-3^3=0$$

Commented by pete last updated on 17/May/20

$$\mathrm{Thanks}\mathrm{very}\mathrm{much}\mathrm{senior}.$$

Commented by niroj last updated on 17/May/20

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Answered by Ar Brandon last updated on 17/May/20

$${\left(3\mathrm{y}\right)}^2+{\left(\frac{1}{\mathrm{y}}\right)}^2={(3\mathrm{y}+\frac{1}{\mathrm{y}})}^2-6=3$$$$\Rightarrow 3\mathrm{y}+\frac{1}{\mathrm{y}}=\pm 3$$$$\Rightarrow {27\mathrm{y}}^3+\frac{1}{{\mathrm{y}}^3}=(3\mathrm{y}+\frac{1}{\mathrm{y}})[{(3\mathrm{y}+\frac{1}{\mathrm{y}})}^2-9]=(\pm 3)[{(\pm 3)}^2-9]=0$$

Commented by pete last updated on 17/May/20

$$\mathrm{thanks}\mathrm{sir}$$

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