If-x-1-x-1-find-out-value-x-20-x-17-x-14-x-11-x-17-x-14-x-11-x-8-

Question Number 66058 by arvinddayama01@gmail.comm last updated on 08/Aug/19

I f x + \frac{1}{x} = 1

f i n d o u t v a l u e : -

\frac{x^{20} + x^{17} + x^{14} + x^{11}}{x^{17} + x^{14} + x^{11} + x^{8}} = ?

Commented by Prithwish sen last updated on 08/Aug/19

x^{2} - x + 1 = 0 \Rightarrow (x + 1) (x^{2} - x + 1) = 0 \Rightarrow x^{3} = - 1

\frac{x^{3} (x^{17} + x^{14} + x^{11} + x^{8})}{x^{17} + x^{14} + x^{11} + x^{8}} = x^{3} = - 1

Commented by $@ty@m123 last updated on 08/Aug/19

I h a v e a d o u b t :

I f y o u w r i t e

x^{2} - x + 1 = 0 \Rightarrow (x + 1) (x^{2} - x + 1) = 0 \Rightarrow x^{3} = - 1

t h e n y o u c a n a l s o w r i t e :

x^{2} - x + 1 = 0 \Rightarrow (x + n) (x^{2} - x + 1) = 0 \Rightarrow x = - n \Rightarrow n \in R

Commented by Prithwish sen last updated on 08/Aug/19

I have just used the formula of

a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2}) where I just

put a = x and b = 1 .

Commented by Prithwish sen last updated on 09/Aug/19

I think the equation x^{3} + 1 = 0 has 3

roots they are - 1, \frac{1 \pm \sqrt{3} i}{2 .}

Now x + 1 = 0 has the root x = - 1

and x^{2} - x + 1 = 0 has two roots \frac{1 \pm \sqrt{3} i}{2}

just like the equation

x^{2} - 5 x + 6 = 0 has 2 roots 2 and 3

of which x - 2 = 0 is satisfied by x = 2

and x - 3 = 0 is satisfied by x = 3 . But as a total

the entire equation is satisfied by both 2

roots .

Commented by $@ty@m123 last updated on 09/Aug/19

O n e m o r e o b s e r v a t i o n :

T h e g i v e n e x p r e s s i o n \neq x^{3}

w h e n x = - 1

Answered by $@ty@m123 last updated on 08/Aug/19

x + \frac{1}{x} = 1

x^{2} - x + 1 = 0

x = \frac{1 \pm \sqrt{3} i}{2}

T h e g i v e n e x p r e s s i o n = x^{3}

= {(\frac{1 \pm \sqrt{3} i}{2})}^{3}

= \frac{1}{8} {1 \pm {(\sqrt{3} i)}^{3} \pm 3 \sqrt{3} i + 3 . (- 3)}

= \frac{1}{8} (1 \mp 3 \sqrt{3} i \mp 3 \sqrt{3} i - 9)

= \frac{- 8}{8}

= - 1

Answered by afjyormathchamp@gmail.com last updated on 08/Aug/19

∴ x^{2} - x + 1 = 0

(x + 1) (x^{2} - x + 1) = 0 x^{3} = - 1

\Rightarrow \frac{x^{3} (x^{17} + x^{14} + x^{11} + x^{8})}{(x^{17} + x^{14} + x^{11} + x^{8})} = x^{3} = - 1