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Question Number 173997 by mnjuly1970 last updated on 22/Jul/22

                If                    (x)^(1/3)  − 3 = ((x−36))^(1/3)                then                       ((x^( 2) −1)/x) = ?

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{If}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sqrt[{\mathrm{3}}]{{x}}\:−\:\mathrm{3}\:=\:\sqrt[{\mathrm{3}}]{{x}−\mathrm{36}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{then}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{x}^{\:\mathrm{2}} −\mathrm{1}}{{x}}\:=\:?\: \\ $$$$\:\:\:\:\:\:\: \\ $$

Answered by behi834171 last updated on 23/Jul/22

x=t^3 ,x−36=s^3   ⇒ { ((t−s=3     ⇒(t−s)[(t−s)^2 +3ts]=36⇒)),((t^3 −s^3 =36)) :}     3(9+3ts)=36⇒ts=1⇒t+s=±[(t−s)^2 +4ts]^(1/2) =  ⇒t+s=±(√(13))  ⇒z^2 −(±(√(13)))z+1=0  ⇒z=((±(√(13))±3)/2)⇒ { ((s=((±(√(13))+3)/2))),((t=((±(√(13))−3)/2))) :}  x=t^3 =(((±(√(13))+3)/2))^3 = { (((13(√(13))+117+27(√(13))+27)/8)),(((−13(√(13))+117−27(√(13))+27)/8)) :}  = { ((5(√(13))+18)),((−5(√(13))+18)) :}  ⇒((x^2 −1)/x)=((25×13+180(√(13))+364−1)/(5(√(13))+18))=((180(√(13))+648)/(5(√(13))+18))     =((36(5(√(13))+18))/(5(√(13))+18))=36     .■

$${x}={t}^{\mathrm{3}} ,{x}−\mathrm{36}={s}^{\mathrm{3}} \\ $$$$\Rightarrow\begin{cases}{{t}−{s}=\mathrm{3}\:\:\:\:\:\Rightarrow\left({t}−{s}\right)\left[\left({t}−{s}\right)^{\mathrm{2}} +\mathrm{3}{ts}\right]=\mathrm{36}\Rightarrow}\\{{t}^{\mathrm{3}} −{s}^{\mathrm{3}} =\mathrm{36}}\end{cases} \\ $$$$\:\:\:\mathrm{3}\left(\mathrm{9}+\mathrm{3}{ts}\right)=\mathrm{36}\Rightarrow{ts}=\mathrm{1}\Rightarrow{t}+{s}=\pm\left[\left({t}−{s}\right)^{\mathrm{2}} +\mathrm{4}{ts}\right]^{\frac{\mathrm{1}}{\mathrm{2}}} = \\ $$$$\Rightarrow{t}+{s}=\pm\sqrt{\mathrm{13}} \\ $$$$\Rightarrow{z}^{\mathrm{2}} −\left(\pm\sqrt{\mathrm{13}}\right){z}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\boldsymbol{\mathrm{z}}=\frac{\pm\sqrt{\mathrm{13}}\pm\mathrm{3}}{\mathrm{2}}\Rightarrow\begin{cases}{{s}=\frac{\pm\sqrt{\mathrm{13}}+\mathrm{3}}{\mathrm{2}}}\\{{t}=\frac{\pm\sqrt{\mathrm{13}}−\mathrm{3}}{\mathrm{2}}}\end{cases} \\ $$$${x}={t}^{\mathrm{3}} =\left(\frac{\pm\sqrt{\mathrm{13}}+\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{3}} =\begin{cases}{\frac{\mathrm{13}\sqrt{\mathrm{13}}+\mathrm{117}+\mathrm{27}\sqrt{\mathrm{13}}+\mathrm{27}}{\mathrm{8}}}\\{\frac{−\mathrm{13}\sqrt{\mathrm{13}}+\mathrm{117}−\mathrm{27}\sqrt{\mathrm{13}}+\mathrm{27}}{\mathrm{8}}}\end{cases} \\ $$$$=\begin{cases}{\mathrm{5}\sqrt{\mathrm{13}}+\mathrm{18}}\\{−\mathrm{5}\sqrt{\mathrm{13}}+\mathrm{18}}\end{cases} \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}}=\frac{\mathrm{25}×\mathrm{13}+\mathrm{180}\sqrt{\mathrm{13}}+\mathrm{364}−\mathrm{1}}{\mathrm{5}\sqrt{\mathrm{13}}+\mathrm{18}}=\frac{\mathrm{180}\sqrt{\mathrm{13}}+\mathrm{648}}{\mathrm{5}\sqrt{\mathrm{13}}+\mathrm{18}} \\ $$$$\:\:\:=\frac{\mathrm{36}\left(\mathrm{5}\sqrt{\mathrm{13}}+\mathrm{18}\right)}{\mathrm{5}\sqrt{\mathrm{13}}+\mathrm{18}}=\mathrm{36}\:\:\:\:\:.\blacksquare \\ $$

Commented by Tawa11 last updated on 23/Jul/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 22/Jul/22

x−27−9(x)^(1/3) ((x)^(1/3) −3)=x−36  (x)^(1/3) ((x)^(1/3) −3)=1  (x)^(1/3) −(1/( (x)^(1/3) ))=3  x−(1/x)−3×3=27  x−(1/x)=36=((x^2 −1)/x)

$${x}−\mathrm{27}−\mathrm{9}\sqrt[{\mathrm{3}}]{{x}}\left(\sqrt[{\mathrm{3}}]{{x}}−\mathrm{3}\right)={x}−\mathrm{36} \\ $$$$\sqrt[{\mathrm{3}}]{{x}}\left(\sqrt[{\mathrm{3}}]{{x}}−\mathrm{3}\right)=\mathrm{1} \\ $$$$\sqrt[{\mathrm{3}}]{{x}}−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{x}}}=\mathrm{3} \\ $$$${x}−\frac{\mathrm{1}}{{x}}−\mathrm{3}×\mathrm{3}=\mathrm{27} \\ $$$${x}−\frac{\mathrm{1}}{{x}}=\mathrm{36}=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}} \\ $$

Commented by Tawa11 last updated on 23/Jul/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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