Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 161169 by MathsFan last updated on 13/Dec/21

if 9x^2 +(1/x^2 )=3 then 27x^3 +(1/x^3 )=?

$${if}\:\mathrm{9}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{3} \\ $$$${then} \\ $$$$\:\mathrm{27}{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=? \\ $$

Answered by Rasheed.Sindhi last updated on 13/Dec/21

9x^2 +(1/x^2 )=3 ; 27x^3 +(1/x^3 )=? (3x+(1/x))^2 =9x^2 +6+(1/x^2 )=3+6=9 3x+(1/x)=±3 (3x+(1/x))^3 =(±3)^3 27x^3 +(1/x^3 )+3(3x)((1/x))(3x+(1/x))=±27 27x^3 +(1/x^3 )+9(±3)=±27 27x^3 +(1/x^3 )±27=±27 27x^3 +(1/x^3 )=±27∓27=0

$$\mathrm{9}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{3}\:;\:\:\mathrm{27}{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=? \\ $$$$\left(\mathrm{3}{x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{9}{x}^{\mathrm{2}} +\mathrm{6}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{3}+\mathrm{6}=\mathrm{9} \\ $$$$\:\mathrm{3}{x}+\frac{\mathrm{1}}{{x}}=\pm\mathrm{3} \\ $$$$\:\left(\mathrm{3}{x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} =\left(\pm\mathrm{3}\right)^{\mathrm{3}} \\ $$$$\mathrm{27}{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\mathrm{3}\left(\mathrm{3}{x}\right)\left(\frac{\mathrm{1}}{{x}}\right)\left(\mathrm{3}{x}+\frac{\mathrm{1}}{{x}}\right)=\pm\mathrm{27} \\ $$$$\mathrm{27}{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\mathrm{9}\left(\pm\mathrm{3}\right)=\pm\mathrm{27} \\ $$$$\mathrm{27}{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\pm\mathrm{27}=\pm\mathrm{27} \\ $$$$\mathrm{27}{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\pm\mathrm{27}\mp\mathrm{27}=\mathrm{0} \\ $$

Answered by Rasheed.Sindhi last updated on 15/Dec/21

27x^3 +(1/x^3 )=(3x+(1/x))(9x^2 −3+(1/x^2 )) =(3x+(1/x))(3−3) =(3x+(1/x))(0)=0 USED FORMULA: determinant (((a^3 +b^3 =(a+b)(a^2 −ab+b^2 ))))

$$\mathrm{27}{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left(\mathrm{3}{x}+\frac{\mathrm{1}}{{x}}\right)\left(\mathrm{9}{x}^{\mathrm{2}} −\mathrm{3}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{3}{x}+\frac{\mathrm{1}}{{x}}\right)\left(\mathrm{3}−\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{3}{x}+\frac{\mathrm{1}}{{x}}\right)\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{USED}\:\mathrm{FORMULA}: \\ $$$$\:\:\:\:\:\:\:\:\begin{array}{|c|}{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\left({a}+{b}\right)\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right)}\\\hline\end{array}\: \\ $$

Commented by MathsFan last updated on 13/Dec/21

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com