solve-for-x-R-x-3-3-x-1-3-3-0-

Question Number 170389 by mr W last updated on 22/May/22

s o l v e f o r x \in R

x^{3} + \sqrt[3]{3 - x} - 3 = 0

Commented by Mastermind last updated on 22/May/22

W e c a n a c t u a l l y s o l v e t h i s b y g r a p h

m e t h o d

O k a y, G r a p h D e t a i l s

y = 3 - x^{3}

y = \sqrt[3]{3 - x}

R o o t (\sqrt[3]{3}, 0)

D o m a i n x \in R

R a n g e y \in R

V e r t i c a l i n t e r c e p t (0, 3)

R o o t (3, 0)

D o m a i n x \in R

R a n g e y \in R

V e r t i c a l i n t e r c e p t (0, \sqrt[3]{3})

Commented by mr W last updated on 22/May/22

w h a t i s y o u r a n s w e r ?

Commented by Mastermind last updated on 22/May/22

y o u h a v e a r r a n g e d i t a g a i n

Commented by jasem1994hamoud last updated on 22/May/22

b y w h a t s a p

+ 79951154287

Commented by mr W last updated on 22/May/22

i j u s t p u t a l l t e r m s o n o n e s i d e . t h e

e q u a t i o n i s t h e s a m e a s b e f o r e .

Commented by Mastermind last updated on 22/May/22

y o u c a n a c t u a l l y a d d m e t o o

+ 2347068402491

Commented by Mastermind last updated on 22/May/22

I u n d e r s t a n d

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

3 - x^{3} = \sqrt[3]{3 - x}

f (x) = 3 - x^{3} \Rightarrow f^{- 1} (x) = \sqrt[3]{3 - x}

\Rightarrow f (x) = f^{- 1} (x)

\Rightarrow f (x) = f^{- 1} (x) = x

3 - x^{3} = x o r \sqrt[3]{3 - x} = x (t h e s a m e)

x^{3} + x - 3 = 0

\Rightarrow x = \sqrt[3]{\sqrt{{(\frac{1}{3})}^{3} + {(- \frac{3}{2})}^{2}} + \frac{3}{2}} - \sqrt[3]{\sqrt{{(\frac{1}{3})}^{3} + {(- \frac{3}{2})}^{2}} - \frac{3}{2}}

\Rightarrow x = \sqrt[3]{\frac{\sqrt{741}}{18} + \frac{3}{2}} - \sqrt[3]{\frac{\sqrt{741}}{18} - \frac{3}{2}} ✓

Commented by Mastermind last updated on 22/May/22

W o w

{t h a t}^{'} s g o o d

Commented by Rasheed.Sindhi last updated on 23/May/22

G reat S ir!

Commented by MJS_new last updated on 23/May/22

I think the given equation also has got a pair

of complex solutions . can we find the exact

values ?

Commented by mr W last updated on 23/May/22

i^{'} m n o t s u r e i f t h e c o m p l e x r o o t s o f

x^{3} + x - 3 = 0 a r e a l s o t h e c o m p l e x r o o t s

o f t h e o r i g i n a l e q u a t i o n . i g u e s s y e s,

b u t n o t s u r e .

Commented by MJS_new last updated on 23/May/22

no . I aporoximated all the roots

(1)

x^{3} + \sqrt[3]{3 - x} - 3 = 0

x_{1} \approx 1.21341166

x_{2, 3} \approx - . 597562988 \pm . 964934775 i

these roots {don}^{'} t give a “ {nice}^{″} polynome :

x^{3} - . 0182856868 x^{2} - 1.56309342

(2)

x^{3} + x - 3 = 0

x_{1} \approx 1.21341166

x_{2, 3} \approx - . 606705831 \pm 1 . 45061225 i

(3)

{[{(3 - x)}^{1 / 3} = 3 - x^{3}]}^{3}

\Rightarrow

x^{9} - 9 x^{6} + 27 x^{3} - x - 24 = 0

(x^{3} + x - 3) (x^{6} - x^{4} - 6 x^{3} + x^{2} + 3 x + 8) = 0

x^{6} - 4^{4} + 6 x^{3} + x^{2} + 3 x + 8 = 0

x_{1, 2} \approx 1.54420759 \pm . 135231222 i

x_{3, 4} \approx - . 597562988 \pm . 964934775 i

x_{5, 6} \approx - . 946644598 \pm 1 . 29938754 i

Commented by mr W last updated on 23/May/22

t h a n k s f o r t h i s i n t e r e s t i n g t h i n g!

Commented by Tawa11 last updated on 08/Oct/22

Great sir

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