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


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

$I f$ $\sqrt[3]{x} - 3 = \sqrt[3]{x - 36}$ $t h e n$ $\frac{x^{2} - 1}{x} = ?$
	Answered by behi834171 last updated on 23/Jul/22
	$x=t^3 ,x−36=s^3 ⇒ { ((t−s=3 ⇒(t−s)[(t−s)^2 +3ts]=36⇒)),((t^3 −s^3 =36)) :} 3(9+3ts)=36⇒ts=1⇒t+s=±[(t−s)^2 +4ts]^(1/2) = ⇒t+s=±(√(13)) ⇒z^2 −(±(√(13)))z+1=0 ⇒z=((±(√(13))±3)/2)⇒ { ((s=((±(√(13))+3)/2))),((t=((±(√(13))−3)/2))) :} x=t^3 =(((±(√(13))+3)/2))^3 = { (((13(√(13))+117+27(√(13))+27)/8)),(((−13(√(13))+117−27(√(13))+27)/8)) :} = { ((5(√(13))+18)),((−5(√(13))+18)) :} ⇒((x^2 −1)/x)=((25×13+180(√(13))+364−1)/(5(√(13))+18))=((180(√(13))+648)/(5(√(13))+18)) =((36(5(√(13))+18))/(5(√(13))+18))=36 .■$
	$x = t^{3}, x - 36 = s^{3}$ $\Rightarrow {\begin{cases} t - s = 3 \Rightarrow (t - s) [{(t - s)}^{2} + 3 t s] = 36 \Rightarrow \\ t^{3} - s^{3} = 36 \end{cases}$ $3 (9 + 3 t s) = 36 \Rightarrow t s = 1 \Rightarrow t + s = \pm {[{(t - s)}^{2} + 4 t s]}^{\frac{1}{2}} =$ $\Rightarrow t + s = \pm \sqrt{13}$ $\Rightarrow z^{2} - (\pm \sqrt{13}) z + 1 = 0$ $\Rightarrow z = \frac{\pm \sqrt{13} \pm 3}{2} \Rightarrow {\begin{cases} s = \frac{\pm \sqrt{13} + 3}{2} \\ t = \frac{\pm \sqrt{13} - 3}{2} \end{cases}$ $x = t^{3} = {(\frac{\pm \sqrt{13} + 3}{2})}^{3} = {\begin{cases} \frac{13 \sqrt{13} + 117 + 27 \sqrt{13} + 27}{8} \\ \frac{- 13 \sqrt{13} + 117 - 27 \sqrt{13} + 27}{8} \end{cases}$ $= {\begin{cases} 5 \sqrt{13} + 18 \\ - 5 \sqrt{13} + 18 \end{cases}$ $\Rightarrow \frac{x^{2} - 1}{x} = \frac{25 \times 13 + 180 \sqrt{13} + 364 - 1}{5 \sqrt{13} + 18} = \frac{180 \sqrt{13} + 648}{5 \sqrt{13} + 18}$ $= \frac{36 (5 \sqrt{13} + 18)}{5 \sqrt{13} + 18} = 36 . ◼$
	Commented by Tawa11 last updated on 23/Jul/22

	$Great sir$
	Answered by mr W last updated on 22/Jul/22

	$x - 27 - 9 \sqrt[3]{x} (\sqrt[3]{x} - 3) = x - 36$ $\sqrt[3]{x} (\sqrt[3]{x} - 3) = 1$ $\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}} = 3$ $x - \frac{1}{x} - 3 \times 3 = 27$ $x - \frac{1}{x} = 36 = \frac{x^{2} - 1}{x}$
	Commented by Tawa11 last updated on 23/Jul/22

	$Great sir$
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