if-a-10-a-5-1-0-then-a-2005-1-a-2005-a-a-10-a-11-b-a-10-a-5-c-3-a-10-a-5-d-0-with-steps-

Question Number 94521 by Abdulrahman last updated on 19/May/20

if a^{10} + a^{5} + 1 = 0

then a^{2005} + \frac{1}{a^{2005}} = ?

a : a^{10} + a^{11} b : a^{10} + a^{5} c : 3 (a^{10} + a^{5}) d : 0

with steps ?

Commented by peter frank last updated on 19/May/20

c h e c k 92839

Commented by Abdulrahman last updated on 19/May/20

thanks alot

Commented by i jagooll last updated on 19/May/20

answer B

Answered by Abdulrahman last updated on 19/May/20

please solve

Answered by i jagooll last updated on 19/May/20

Commented by i jagooll last updated on 19/May/20

this is mr john's answer

Answered by mathmax by abdo last updated on 20/May/20

1 + a^{5} + a^{10} = 0 \Rightarrow 1 + a^{5} + {(a^{5})}^{2} = 0 \Rightarrow \frac{1 - {(a^{5})}^{3}}{1 - a^{5}} = 0 \Rightarrow a^{15} = 1 and a^{5} \neq 0

the roots of a^{15} = 0 are z_{k} = e^{i \frac{2 k π}{15}} k \in [[0, 14]]

a^{2005} + a^{- 2005} = (e^{i \frac{2 k π (2005)}{15}} + e^{- i \frac{2 k π (2005)}{15}})

= 2 \cos (\frac{2 k π (2005)}{15}) = 2 \cos (\frac{4010 k π}{15}) = 2 \cos (267 π + \frac{π}{3})

= 2 \cos (π + \frac{π}{3}) = - 2 \cos (\frac{π}{3}) = - 2 \times \frac{1}{2} = - 1