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Question Number 155259 by Fridunatjan08 last updated on 27/Sep/21

If: x+(1/x)=6 then x^3 +(1/x^3 )=?

$${If}:\:{x}+\frac{\mathrm{1}}{{x}}=\mathrm{6}\:\:{then}\:\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=? \\ $$

Commented by puissant last updated on 27/Sep/21

(x+(1/x))^3 = 6^3 = 216 (x+(1/x))^3 = x^3 + 3x^2 ×(1/x) + 3x×(1/x^2 ) +(1/x^3 ) =x^3 + (1/x^3 ) + 3(x+(1/x))= 216 ⇒ x^3 +(1/x^3 ) = 216−3×6 = 198..

$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} =\:\mathrm{6}^{\mathrm{3}} \:=\:\mathrm{216} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} =\:{x}^{\mathrm{3}} \:+\:\mathrm{3}{x}^{\mathrm{2}} ×\frac{\mathrm{1}}{{x}}\:+\:\mathrm{3}{x}×\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{x}^{\mathrm{3}} } \\ $$$$={x}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:+\:\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right)=\:\mathrm{216} \\ $$$$\Rightarrow\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:=\:\mathrm{216}−\mathrm{3}×\mathrm{6}\:=\:\mathrm{198}.. \\ $$

Commented by mathdanisur last updated on 27/Sep/21

x + (1/x) = a ⇒ x^3 + (1/x^3 ) = a^3 - 3a ⇒ 198

$$\mathrm{x}\:+\:\frac{\mathrm{1}}{\mathrm{x}}\:=\:\mathrm{a}\:\Rightarrow\:\mathrm{x}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\:=\:\mathrm{a}^{\mathrm{3}} \:-\:\mathrm{3a}\:\Rightarrow\:\mathrm{198} \\ $$

Answered by Rasheed.Sindhi last updated on 28/Sep/21

An Alternate Way determinant (((a^3 +b^3 =(a+b)(a^2 −ab+b^2 ))) x^3 +(1/x^3 )=(x+(1/x))(x^2 −1+(1/x^2 )) =(x+(1/x))((x+(1/x))^2 −2−1) =(6)(6^2 −3)=6×33=198

$$\mathbb{A}\mathrm{n}\:\:\mathbb{A}\mathrm{lternate}\:\mathbb{W}\mathrm{ay} \\ $$$$\:\:\:\begin{array}{|c|}{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\left({a}+{b}\right)\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right.}\\\hline\end{array}\:\:\: \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left({x}+\frac{\mathrm{1}}{{x}}\right)\left({x}^{\mathrm{2}} −\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right) \\ $$$$\:\:\:=\left({x}+\frac{\mathrm{1}}{{x}}\right)\left(\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:=\left(\mathrm{6}\right)\left(\mathrm{6}^{\mathrm{2}} −\mathrm{3}\right)=\mathrm{6}×\mathrm{33}=\mathrm{198}\:\:\:\:\:\:\:\: \\ $$

Commented by peter frank last updated on 28/Sep/21

great

$$\mathrm{great} \\ $$

Commented by Rasheed.Sindhi last updated on 28/Sep/21

Thank you!

$$\mathrm{Thank}\:\mathrm{you}! \\ $$

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