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

If 9y^2 + (1/y^2 ) =3, then find the value of  27y^3  + (1/y^3 )

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

Answered by niroj last updated on 17/May/20

  (3y+(1/y))^2 −2.3y.(1/y)=9y^2 +(1/y^2 )      (3y+(1/y))^2 =3+6         3y+(1/y)=3    now,    (3y+(1/y))^3 = (3y)^3 +((1/y))^3 +3.3y.(1/y)(3y+(1/y))                       = 27y^3 +(1/y^3 )+9.3   27y^3 +(1/y^3 )= (3y+(1/y))^3 −27           = 3^3 −3^3 =0

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

Commented by pete last updated on 17/May/20

Thanks very much senior.

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

Commented by niroj last updated on 17/May/20

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

(3y)^2 +((1/y))^2 =(3y+(1/y))^2 −6=3  ⇒3y+(1/y)=±3  ⇒27y^3 +(1/y^3 )=(3y+(1/y))[(3y+(1/y))^2 −9]=(±3)[(±3)^2 −9]=0

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

Commented by pete last updated on 17/May/20

thanks sir

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

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