Question Number 223429 by klipto last updated on 25/Jul/25
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Commented by klipto last updated on 25/Jul/25
Answered by Rasheed.Sindhi last updated on 26/Jul/25
$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{51},\:{x}^{\mathrm{3}} −\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=? \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{2}=\mathrm{51}−\mathrm{2}=\mathrm{49} \\ $$$$\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{49} \\ $$$${x}−\frac{\mathrm{1}}{{x}}=\pm\mathrm{7} \\ $$$$\: \\ $$$${x}^{\mathrm{3}} −\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left({x}−\frac{\mathrm{1}}{{x}}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}\right) \\ $$$${x}^{\mathrm{3}} −\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left(\pm\mathrm{7}\right)\left(\mathrm{51}+\mathrm{1}\right) \\ $$$${x}^{\mathrm{3}} −\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\pm\mathrm{364} \\ $$
Commented by Rasheed.Sindhi last updated on 25/Jul/25
$${Thanks}\:\:{sir},\:{I}'{ll}\:{go}\:{to}\:{correct}! \\ $$
Commented by Ghisom last updated on 25/Jul/25
$${x}^{\mathrm{3}} −\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left({x}−\frac{\mathrm{1}}{{x}}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}\right) \\ $$$${x}^{\mathrm{3}} −\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left(\pm\mathrm{7}\right)\left(\mathrm{51}+\mathrm{1}\right)=\pm\mathrm{364} \\ $$
Answered by fantastic last updated on 25/Jul/25
![x^2 +(1/x^2 )=51 or (x)^2 +((1/x))^2 =51 or (x−(1/x))^2 +2×x×(1/x)=51 So (x−(1/x))^2 =51−2=49 So (x−(1/x))=(√(49))=7[all options are positive] x^3 −(1/x^3 )=(x−(1/x))^3 +3×x×(1/x)(x+(1/x)) =7^3 +3×7=343+21=364 So option (a)364✓](https://www.tinkutara.com/question/Q223433.png)
$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{51} \\ $$$${or}\:\left({x}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{51} \\ $$$${or}\:\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}×\cancel{{x}}×\frac{\mathrm{1}}{\cancel{{x}}}=\mathrm{51} \\ $$$${So}\:\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{51}−\mathrm{2}=\mathrm{49} \\ $$$${So}\:\left({x}−\frac{\mathrm{1}}{{x}}\right)=\sqrt{\mathrm{49}}=\mathrm{7}\left[{all}\:{options}\:{are}\:{positive}\right] \\ $$$${x}^{\mathrm{3}} −\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} +\mathrm{3}×{x}×\frac{\mathrm{1}}{{x}}\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$=\mathrm{7}^{\mathrm{3}} +\mathrm{3}×\mathrm{7}=\mathrm{343}+\mathrm{21}=\mathrm{364}\: \\ $$$${So}\:{option}\:\left({a}\right)\mathrm{364}\checkmark \\ $$