if-x-1-x-3-find-x-24-x-18-x-6-1-

Question Number 60905 by necx1 last updated on 27/May/19

i f x + \frac{1}{x} = \sqrt{3} . f i n d

x^{24} + x^{18} + x^{6} + 1

Answered by ajfour last updated on 27/May/19

E = x^{24} + x^{18} + x^{6} + 1

x^{2} - \sqrt{3} x + 1 = 0

x = \frac{\sqrt{3} \pm i}{2} = e^{\pm i \frac{π}{6}}

E = 1 - 1 - 1 + 1 = 0 .

Commented by necx1 last updated on 27/May/19

t h a n k y o u s i r

Answered by Rasheed.Sindhi last updated on 27/May/19

x + \frac{1}{x} = \sqrt{3} (G i v e n)

{(x + \frac{1}{x})}^{2} = {(\sqrt{3})}^{2}

x^{2} + \frac{1}{x^{2}} = 3 - 2 = 1

{(x^{2} + \frac{1}{x^{2}})}^{3} = {(1)}^{3} = 1

{(x^{2})}^{3} + \frac{1}{{(x^{2})}^{3}} + 3 . x^{2} . \frac{1}{x^{2}} (x^{2} + \frac{1}{x^{2}}) = 1

x^{6} + \frac{1}{x^{6}} = 1 - 3 = - 2 \dots \dots \dots \dots \dots \dots \dots \dots A

{(x^{6} + \frac{1}{x^{6}})}^{2} = {(- 2)}^{2} = 4

x^{12} + \frac{1}{x^{12}} = 4 - 2 = 2 \dots \dots \dots \dots \dots \dots \dots . B

\Rightarrow x^{24} - 2 x^{12} + 1 = 0

\Rightarrow {(x^{12} - 1)}^{2} = 0 \Rightarrow x^{12} = 1 \dots \dots \dots . . C

N o w,

L e t P = x^{24} + x^{18} + x^{6} + 1

P = x^{12} (x^{12} + \frac{1}{x^{12}} + x^{6} + \frac{1}{x^{6}})

= (1) (2 - 2) = 0

Commented by necx1 last updated on 27/May/19

t o o m a n y m e t h o d s f o r a p p r o a c h i n g t h e

p r o b l e m . I^{'} m s o g r a t e f u l .

Answered by math1967 last updated on 27/May/19

x + \frac{1}{x} = \sqrt{3} \Rightarrow x^{3} + \frac{1}{x^{3}} = 0 \frac{x^{6} + 1}{x^{3}} = 0

∴ x^{6} + 1 = 0 ∴ x^{24} + x^{18} + x^{6} + 1 = x^{18} (x^{6} + 1) + x^{6} + 1

= x^{18} \times 0 + 0 = 0 a n s

Commented by Rasheed.Sindhi last updated on 27/May/19

M o r e e f f i c i e n t a p p r o a c h t h a n m i n e!

Commented by math1967 last updated on 27/May/19

T h a n k y o u s i r