((○BeMath○)/(∧⌣∧)) { ((x^4 +(1/x^4 ) = 23)),((x^3 −(1/x^3 ) = ?)) :}


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Question Number 107945 by bemath last updated on 13/Aug/20
$((○BeMath○)/(∧⌣∧)) { ((x^4 +(1/x^4 ) = 23)),((x^3 −(1/x^3 ) = ?)) :}$
$\frac{\circ B e M a t h \circ}{\land ⌣ \land}$ ${\begin{cases} x^{4} + \frac{1}{x^{4}} = 23 \\ x^{3} - \frac{1}{x^{3}} = ? \end{cases}$
	Answered by $@y@m last updated on 13/Aug/20

	${(x^{2} + \frac{1}{x^{2}})}^{2} - 2 = 23$ ${(x^{2} + \frac{1}{x^{2}})}^{2} = 25$ $x^{2} + \frac{1}{x^{2}} = 5$ $x^{2} + \frac{1}{x^{2}} - 2 = 3$ ${(x - \frac{1}{x})}^{2} = 3$ $x - \frac{1}{x} = \sqrt{3}$ ${(x - \frac{1}{x})}^{3} = 3 \sqrt{3}$ $x^{3} - \frac{1}{x^{3}} - 3 (x - \frac{1}{x}) = 3 \sqrt{3}$ $x^{3} - \frac{1}{x^{3}} - 3 \sqrt{3} = 3 \sqrt{3}$ $x^{3} - \frac{1}{x^{3}} = 6 \sqrt{3}$
	Commented by Rasheed.Sindhi last updated on 13/Aug/20

	$Nic e!$
	Commented by Her_Majesty last updated on 13/Aug/20

	$n i c e b u t a l s o - 6 \sqrt{3} i s a s o l u t i o n$
	Commented by bemath last updated on 13/Aug/20

	$y e s s i r . t h e a n s w e r \pm 6 \sqrt{3}$
	Answered by 1549442205PVT last updated on 13/Aug/20
	$x^4 +(1/x^4 ) = 23⇔(x^2 +(1/x^2 ))^2 =25 ⇔x^2 +(1/x^2 )=±5 i)x^2 +(1/x^2 )=5⇔(x−(1/x))^2 =3⇔x−(1/x)=±(√3) a)x−(1/x)=(√3) ⇒x^3 −(1/x^3 )=(x−(1/x))^3 +3(x−(1/x)) =3(√3)+3(√3)=6(√3) b)x−(1/x)=−(√3)⇒x^3 −(1/x^3 )=(−(√3))^3 +3.(−(√3)) =−6(√3) ii)x^2 +(1/x^2 )=−5⇒(x−(1/x))^2 =−7 ⇒x−(1/x)=±i(√7) c)x−(1/x)=i(√7)⇒x^3 −(1/x^3 )=(i(√7))^3 +3i(√7) =−7i(√7)+3i(√7)=−4i(√7) d)x−(1/x)=−i(√7)⇒x^3 −(1/x^3 )=(−i(√7))^3 −3i(√7) =7i(√7)−3i(√7)=4i(√7) Thus,we get four answers follows as: x^3 −(1/x^3 )∈{6(√3);−6(√3);4i(√7);−4i(√7)}$
	$x^{4} + \frac{1}{x^{4}} = 23 \Leftrightarrow {(x^{2} + \frac{1}{x^{2}})}^{2} = 25$ $\Leftrightarrow x^{2} + \frac{1}{x^{2}} = \pm 5$ $i) x^{2} + \frac{1}{x^{2}} = 5 \Leftrightarrow {(x - \frac{1}{x})}^{2} = 3 \Leftrightarrow x - \frac{1}{x} = \pm \sqrt{3}$ $a) x - \frac{1}{x} = \sqrt{3} \Rightarrow x^{3} - \frac{1}{x^{3}} = {(x - \frac{1}{x})}^{3} + 3 (x - \frac{1}{x})$ $= 3 \sqrt{3} + 3 \sqrt{3} = 6 \sqrt{3}$ $b) x - \frac{1}{x} = - \sqrt{3} \Rightarrow x^{3} - \frac{1}{x^{3}} = {(- \sqrt{3})}^{3} + 3 . (- \sqrt{3})$ $= - 6 \sqrt{3}$ $ii) x^{2} + \frac{1}{x^{2}} = - 5 \Rightarrow {(x - \frac{1}{x})}^{2} = - 7$ $\Rightarrow x - \frac{1}{x} = \pm i \sqrt{7}$ $c) x - \frac{1}{x} = i \sqrt{7} \Rightarrow x^{3} - \frac{1}{x^{3}} = {(i \sqrt{7})}^{3} + 3 i \sqrt{7}$ $= - 7 i \sqrt{7} + 3 i \sqrt{7} = - 4 i \sqrt{7}$ $d) x - \frac{1}{x} = - i \sqrt{7} \Rightarrow x^{3} - \frac{1}{x^{3}} = {(- i \sqrt{7})}^{3} - 3 i \sqrt{7}$ $= 7 i \sqrt{7} - 3 i \sqrt{7} = 4 i \sqrt{7}$ $Thus, we get four answers follows as :$ $x^{3} - \frac{1}{x^{3}} \in {6 \sqrt{3}; - 6 \sqrt{3}; 4 i \sqrt{7}; - 4 i \sqrt{7}}$
	Answered by Rasheed.Sindhi last updated on 13/Aug/20
	$An uncommon way_↘ ^↗ → { ((x^4 +(1/x^4 ) = 23)),((x^3 −(1/x^3 ) = ?)) :} (x^4 +(1/x^4 ))^3 =(23)^3 x^(12) +(1/x^(12) )+3(x^4 +(1/x^4 ))=23^3 x^(12) +(1/x^(12) )=23^3 −3(23)=12098 (x^6 +(1/x^6 ))^2 −2=12098 (x^6 +(1/x^6 ))^2 =12100 x^6 +(1/x^6 )=±110 x^6 +(1/x^6 )−2=±110−2 (x^3 −(1/x^3 ))^2 =108,112 x^3 −(1/x^3 )=±6(√3) ,±4(√7) ⟨⋊_↙ ^↖ sindhi_↘ ^↗ ⋉⟩_(↓) ^(↑) _ ^$
	$An uncommon {way}_{↘}^{↗} \to$ ${\begin{cases} x^{4} + \frac{1}{x^{4}} = 23 \\ x^{3} - \frac{1}{x^{3}} = ? \end{cases}$ ${(x^{4} + \frac{1}{x^{4}})}^{3} = {(23)}^{3}$ $x^{12} + \frac{1}{x^{12}} + 3 (x^{4} + \frac{1}{x^{4}}) = 23^{3}$ $x^{12} + \frac{1}{x^{12}} = 23^{3} - 3 (23) = 12098$ ${(x^{6} + \frac{1}{x^{6}})}^{2} - 2 = 12098$ ${(x^{6} + \frac{1}{x^{6}})}^{2} = 12100$ $x^{6} + \frac{1}{x^{6}} = \pm 110$ $x^{6} + \frac{1}{x^{6}} - 2 = \pm 110 - 2$ ${(x^{3} - \frac{1}{x^{3}})}^{2} = 108, 112$ $x^{3} - \frac{1}{x^{3}} = \pm 6 \sqrt{3}, \pm 4 \sqrt{7}$ $\underset{↓}{\overset{↑}{⟨ ⋊_{↙}^{↖} {sindhi}_{↘}^{↗} ⋉ ⟩}}_{}^{}$
	Commented by bemath last updated on 13/Aug/20

	$c o o l l$
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