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Question Number 154385 by amin96 last updated on 17/Sep/21
if (x)^(1/3) −((x−36))^(1/3) =3 find x−(1/x)
$${if}\:\:\:\sqrt[{\mathrm{3}}]{{x}}−\sqrt[{\mathrm{3}}]{{x}−\mathrm{36}}=\mathrm{3}\:\:\:\:\:\:{find}\:\:\:\:{x}−\frac{\mathrm{1}}{{x}} \\ $$
Answered by amin96 last updated on 18/Sep/21
x=t^3 ⇒ t−((t^3 −36))^(1/3) =3 ⇒ ((t^3 −36))^(1/3) =t−3 t^3 −36=t^3 −9t^2 +27t−27 ⇒ t^2 −3t−1=0 t−(1/t)=3 t^2 +(1/t^2 )=11 ⇒ (t^2 +(1/t^2 ))(t−(1/t))=33 t^3 −t+(1/t)−(1/t^3 )=33 ⇒ t^3 −(1/t^3 )=36 x−(1/x)=36
$${x}={t}^{\mathrm{3}} \:\:\:\Rightarrow\:\:\:{t}−\sqrt[{\mathrm{3}}]{{t}^{\mathrm{3}} −\mathrm{36}}=\mathrm{3}\:\:\Rightarrow\:\:\sqrt[{\mathrm{3}}]{{t}^{\mathrm{3}} −\mathrm{36}}={t}−\mathrm{3} \\ $$$${t}^{\mathrm{3}} −\mathrm{36}={t}^{\mathrm{3}} −\mathrm{9}{t}^{\mathrm{2}} +\mathrm{27}{t}−\mathrm{27}\:\:\Rightarrow\:\:{t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{1}=\mathrm{0} \\ $$$${t}−\frac{\mathrm{1}}{{t}}=\mathrm{3}\:\:\:\:{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }=\mathrm{11}\:\:\Rightarrow\:\:\:\left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)\left({t}−\frac{\mathrm{1}}{{t}}\right)=\mathrm{33} \\ $$$${t}^{\mathrm{3}} −{t}+\frac{\mathrm{1}}{{t}}−\frac{\mathrm{1}}{{t}^{\mathrm{3}} }=\mathrm{33}\:\:\Rightarrow\:\:\:{t}^{\mathrm{3}} −\frac{\mathrm{1}}{{t}^{\mathrm{3}} }=\mathrm{36}\:\:\:\:{x}−\frac{\mathrm{1}}{{x}}=\mathrm{36} \\ $$
Answered by liberty last updated on 18/Sep/21
If (x)^(1/3) −((x−36))^(1/3) = 3 then find x−(1/x). solution: ⇒3+((x−36))^(1/3) −(x)^(1/3) = 0 ⇒27+x−36−x=−3×3((x(x−36)))^(1/3) ⇒−9=−9 ((x(x−36)))^(1/3) ⇒1=x^2 −36x ⇒(1/x)=x−36 ⇒x−(1/x)=36
$$\:{If}\:\sqrt[{\mathrm{3}}]{{x}}−\sqrt[{\mathrm{3}}]{{x}−\mathrm{36}}\:=\:\mathrm{3}\:{then} \\ $$$${find}\:{x}−\frac{\mathrm{1}}{{x}}. \\ $$$${solution}: \\ $$$$\Rightarrow\mathrm{3}+\sqrt[{\mathrm{3}}]{{x}−\mathrm{36}}−\sqrt[{\mathrm{3}}]{{x}}\:=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{27}+{x}−\mathrm{36}−{x}=−\mathrm{3}×\mathrm{3}\sqrt[{\mathrm{3}}]{{x}\left({x}−\mathrm{36}\right)} \\ $$$$\Rightarrow−\mathrm{9}=−\mathrm{9}\:\sqrt[{\mathrm{3}}]{{x}\left({x}−\mathrm{36}\right)}\: \\ $$$$\Rightarrow\mathrm{1}={x}^{\mathrm{2}} −\mathrm{36}{x} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}}={x}−\mathrm{36}\:\Rightarrow{x}−\frac{\mathrm{1}}{{x}}=\mathrm{36} \\ $$

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