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If-x-3-2-2-then-x-6-x-4-x-2-1-x-3-

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Question Number 202102 by MATHEMATICSAM last updated on 20/Dec/23
If x = 3 + 2(√2) then ((x^6 + x^4 + x^2 + 1)/x^3 ) = ?
$$\mathrm{If}\:{x}\:=\:\mathrm{3}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{then}\:\frac{{x}^{\mathrm{6}} \:+\:{x}^{\mathrm{4}} \:+\:{x}^{\mathrm{2}} \:+\:\mathrm{1}}{{x}^{\mathrm{3}} }\:=\:? \\ $$
Answered by AST last updated on 20/Dec/23
x+(1/x)=3+2(√2)+3−2(√2)=6 ⇒x^3 +x+(1/x)+(1/x^3 )=(x+(1/x))^3 −2(x+(1/x))=204
$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}=\mathrm{6} \\ $$$$\Rightarrow{x}^{\mathrm{3}} +{x}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} −\mathrm{2}\left({x}+\frac{\mathrm{1}}{{x}}\right)=\mathrm{204} \\ $$
Commented by necx122 last updated on 20/Dec/23
This is wonderful. Thank you.
$${This}\:{is}\:{wonderful}.\:{Thank}\:{you}. \\ $$
Answered by Rasheed.Sindhi last updated on 21/Dec/23
Another Way If x = 3 + 2(√2) then ((x^6 + x^4 + x^2 + 1)/x^3 ) = ? (1/x)=(1/(3+2(√2))) ∙ ((3−2(√2))/(3−2(√2)))=3−2(√2) x+(1/x)=6 ((x^6 + x^4 + x^2 + 1)/x^3 )=((x^4 (x^2 +1)+(x^2 +1))/x^3 ) =(((x^2 +1)(x^4 +1))/x^3 )=((x^2 +1)/x)∙((x^4 +1)/x^2 ) =(x+(1/x))(x^2 +(1/x^2 )) =(x+(1/x))( (x+(1/x))^2 −2) =(6)(6^2 −2)=6(34)=204
$${Another}\:{Way} \\ $$$$\mathrm{If}\:{x}\:=\:\mathrm{3}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{then}\:\frac{{x}^{\mathrm{6}} \:+\:{x}^{\mathrm{4}} \:+\:{x}^{\mathrm{2}} \:+\:\mathrm{1}}{{x}^{\mathrm{3}} }\:=\:? \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{\mathrm{1}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}\:\centerdot\:\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}=\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\: \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{6} \\ $$$$\frac{{x}^{\mathrm{6}} \:+\:{x}^{\mathrm{4}} \:+\:{x}^{\mathrm{2}} \:+\:\mathrm{1}}{{x}^{\mathrm{3}} }=\frac{{x}^{\mathrm{4}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)+\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{{x}^{\mathrm{3}} } \\ $$$$=\frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{4}} +\mathrm{1}\right)}{{x}^{\mathrm{3}} }=\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}}\centerdot\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$=\left({x}+\frac{\mathrm{1}}{{x}}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right) \\ $$$$=\left({x}+\frac{\mathrm{1}}{{x}}\right)\left(\:\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}\right) \\ $$$$=\left(\mathrm{6}\right)\left(\mathrm{6}^{\mathrm{2}} −\mathrm{2}\right)=\mathrm{6}\left(\mathrm{34}\right)=\mathrm{204} \\ $$

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