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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 \dots (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 \dots (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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