an unsolved old question #190875 a, b, c are real roots of the equation x^3 −7x^2 +4x+1=0. find (1/( (a)^(1/3) ))+(1/( (b)^(1/3) ))+(1/( (c)^(1/3) ))=?


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Question Number 195628 by mr W last updated on 06/Aug/23

$You can't use 'macro parameter character #' in math mode$ $a, b, c a r e r e a l r o o t s o f t h e e q u a t i o n$ $x^{3} - 7 x^{2} + 4 x + 1 = 0 .$ $f i n d \frac{1}{\sqrt[3]{a}} + \frac{1}{\sqrt[3]{b}} + \frac{1}{\sqrt[3]{c}} = ?$
Commented by Frix last updated on 06/Aug/23

$I {don}^{'} t think we can give the exact value$
Commented by mr W last updated on 06/Aug/23

$b u t c a n w e e x p r e s s t h e r e s u l t i n a n$ $e x a c t f o r m ? j u s t l i k e \sin \frac{π}{13}, i t i s$ $a n e x a c t f o r m, e v e n w h e n w e {d o n}^{'} t$ $k n o w i t s e x a c t v a l u e .$
Commented by Frix last updated on 06/Aug/23

$In this case (as you certainly know) we get$ $3 trigonometric solutions . I {don}^{'} t think we$ $can simplify Σ \frac{1}{\sqrt[3]{x_{j}}}$ ${We}^{'} d need to find a 3^{rd} degree factor of$ $\frac{1}{y^{9}} - \frac{7}{y^{6}} + \frac{4}{y^{3}} + 1 = 0 \Leftrightarrow y^{9} + 4 y^{6} - 7 y^{3} + 1 = 0$ $with 3 real roots and I {don}^{'} t think {that}^{'} s$ $possible .$
	Answered by witcher3 last updated on 06/Aug/23

	$\frac{1}{\sqrt[3]{a}} . . ?$
	Commented by mr W last updated on 06/Aug/23

	$y e s!$
	Answered by mr W last updated on 06/Aug/23

	$a, b, c a r e r o o t s o f$ $x^{3} - 7 x^{2} + 4 x + 1 = 0 .$ ${(\frac{1}{x})}^{3} + 4 {(\frac{1}{x})}^{2} - 7 (\frac{1}{x}) + 1 = 0$ $s o \frac{1}{a}, \frac{1}{b}, \frac{1}{c} a r e r o o t s o f$ $z^{3} + 4 z^{2} - 7 z + 1 = 0 .$ $l e t p = \frac{1}{a}, q = \frac{1}{b}, r = \frac{1}{c}$ $p + q + r = - 4$ $p q + q r + r p = - 7$ $p q r = - 1$ $s a y s = \frac{1}{\sqrt[3]{a}} + \frac{1}{\sqrt[3]{b}} + \frac{1}{\sqrt[3]{c}} = \sqrt[3]{p} + \sqrt[3]{q} + \sqrt[3]{r}$ $s^{3} = p + q + r - 3 \sqrt[3]{p q r} + 3 s (\sqrt[3]{p q} + \sqrt[3]{q r} + \sqrt[3]{r p})$ $\Rightarrow s^{3} = - 1 + 3 s t . . . (i)$ $w i t h t = \sqrt[3]{p q} + \sqrt[3]{q r} + \sqrt[3]{r p}$ $t^{3} = p q + q r + r p - 3 \sqrt[3]{{(p q r)}^{2}} + 3 t \sqrt[3]{p q r} (\sqrt[3]{p} + \sqrt[3]{q} + \sqrt[3]{r})$ $\Rightarrow t^{3} = - 10 - 3 t s . . . (i i)$ $(i) + (i i) :$ $s^{3} + t^{3} = - 11 \Rightarrow t^{3} = - 11 - s^{3}$ $(i) :$ $s^{3} + 1 = 3 s t$ ${(s^{3} + 1)}^{3} = 27 s^{3} t^{3}$ ${(s^{3} + 1)}^{3} = 27 s^{3} (- 11 - s^{3})$ $l e t u = s^{3}$ ${(u + 1)}^{3} = 27 u (- 11 - u)$ $u^{3} + 30 u^{2} + 300 u + 1 = 0$ $l e t u = v - 10$ ${(v - 10)}^{3} + 30 {(v - 10)}^{2} + 300 (v - 10) + 1 = 0$ $v^{3} - 999 = 0$ $\Rightarrow v = \sqrt[3]{999}$ $\Rightarrow s^{3} = u = v - 10 = \sqrt[3]{999} - 10$ $\Rightarrow s = \sqrt[3]{\sqrt[3]{999} - 10}$ $i . e . \frac{1}{\sqrt[3]{a}} + \frac{1}{\sqrt[3]{b}} + \frac{1}{\sqrt[3]{c}} = - \sqrt[3]{10 - \sqrt[3]{999}} ✓$
	Commented by Frix last updated on 06/Aug/23

	$Nice!$
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