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Question Number 90716 by jagoll last updated on 25/Apr/20

If x + (1/x) = 4 , what the value of ((x^6 −1)/x^3 )

$${If}\:{x}\:+\:\frac{\mathrm{1}}{{x}}\:=\:\mathrm{4}\:,\:{what}\:{the}\: \\ $$$${value}\:{of}\:\frac{{x}^{\mathrm{6}} −\mathrm{1}}{{x}^{\mathrm{3}} } \\ $$

Commented by john santu last updated on 25/Apr/20

x^2 +(1/x^2 ) + 2 = 16 ⇒x^2 +(1/x^2 ) = 14 ((x^6 −1)/x^3 ) = x^3 −(1/x^3 ) = (x−(1/x))(x^2 +(1/x^2 )+1) x^2 +1 = 4x ⇒x^2 −4x+1 = 0 x = ((4 ± 2(√3))/2) = 2 ± (√3) (1) x = 2 + (√( 3)) ⇒(1/x) = 2−(√3) ⇒ x−(1/x) = 2+(√3) − (2−(√3)) = 2(√3) (2) x = 2−(√3) ⇒(1/x) = 2+(√3) ⇒x−(1/x) = 2−(√3) −(2+(√3)) = −2(√3) ∴ ((x^6 −1)/x^3 ) = { ((2(√3) (15) = 30(√3))),((−2(√3) (15) = −30(√3) )) :}

$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\:\mathrm{2}\:=\:\mathrm{16}\:\Rightarrow{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:=\:\mathrm{14} \\ $$$$\frac{{x}^{\mathrm{6}} −\mathrm{1}}{{x}^{\mathrm{3}} }\:=\:{x}^{\mathrm{3}} −\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:=\:\left({x}−\frac{\mathrm{1}}{{x}}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}\right) \\ $$$${x}^{\mathrm{2}} +\mathrm{1}\:=\:\mathrm{4}{x}\:\Rightarrow{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}\:=\:\mathrm{0} \\ $$$${x}\:=\:\frac{\mathrm{4}\:\pm\:\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\:\mathrm{2}\:\pm\:\sqrt{\mathrm{3}} \\ $$$$\left(\mathrm{1}\right)\:{x}\:=\:\mathrm{2}\:+\:\sqrt{\:\mathrm{3}}\:\Rightarrow\frac{\mathrm{1}}{{x}}\:=\:\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\:{x}−\frac{\mathrm{1}}{{x}}\:=\:\mathrm{2}+\sqrt{\mathrm{3}}\:−\:\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\:=\:\mathrm{2}\sqrt{\mathrm{3}}\: \\ $$$$\left(\mathrm{2}\right)\:{x}\:=\:\mathrm{2}−\sqrt{\mathrm{3}}\:\Rightarrow\frac{\mathrm{1}}{{x}}\:=\:\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{x}−\frac{\mathrm{1}}{{x}}\:=\:\mathrm{2}−\sqrt{\mathrm{3}}\:−\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:=\:−\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\therefore\:\frac{{x}^{\mathrm{6}} −\mathrm{1}}{{x}^{\mathrm{3}} }\:=\:\begin{cases}{\mathrm{2}\sqrt{\mathrm{3}}\:\left(\mathrm{15}\right)\:=\:\mathrm{30}\sqrt{\mathrm{3}}}\\{−\mathrm{2}\sqrt{\mathrm{3}}\:\left(\mathrm{15}\right)\:=\:−\mathrm{30}\sqrt{\mathrm{3}}\:}\end{cases} \\ $$

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