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Question Number 26228 by ktomboy1992 last updated on 23/Dec/17

if x^4 +(1/x^4 )=322 find x^3 −(1/x^3 )

$$\mathrm{if}\:\mathrm{x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }=\mathrm{322}\:\mathrm{find}\:\mathrm{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} } \\ $$

Commented by kaivan.ahmadi last updated on 22/Dec/17

x^4 +(1/x^4 )=(x^2 +(1/x^2 ))^2 −2⇒ x^2 +(1/x^2 )=(√(324))=18=(x+(1/x))^2 −2⇒ x+(1/x)=(√(20)) ⇒x^3 +(1/x^3 )=(x+(1/x))^3 −3(x+(1/x))= 20(√(20))−3(√(20))=17(√(20))

$$\mathrm{x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }=\left(\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)^{\mathrm{2}} −\mathrm{2}\Rightarrow \\ $$$$\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\sqrt{\mathrm{324}}=\mathrm{18}=\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} −\mathrm{2}\Rightarrow \\ $$$$\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}=\sqrt{\mathrm{20}} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }=\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)= \\ $$$$\mathrm{20}\sqrt{\mathrm{20}}−\mathrm{3}\sqrt{\mathrm{20}}=\mathrm{17}\sqrt{\mathrm{20}} \\ $$

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