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Question Number 199731 by Rupesh123 last updated on 08/Nov/23

Answered by des_ last updated on 08/Nov/23

(x + (1/x) )^3 = 3(√3) ⇒ ⇒ x^3 + 3(x + (1/x) ) + (1/x^3 ) = 3(√3) ⇒ ⇒ x^3 + (1/x^3 ) = 0 ⇒ x^6 = −1; x^(2023) = (x^6 )^(337) x = −x; x^(2023) + (1/x^(2023) ) = −(x + (1/x)) = −(√3)

(x+1x)3=33⇒⇒x3+3(x+1x)+1x3=33⇒⇒x3+1x3=0⇒x6=−1;x2023=(x6)337x=−x;x2023+1x2023=−(x+1x)=−3

Commented by Rupesh123 last updated on 08/Nov/23

Perfect, sir!

Answered by mathfreak01 last updated on 08/Nov/23

OR x = cos θ + isin θ x + (1/x) = 2cos θ = (√3) θ = cos^(−1) ((√3)/2) = 30° x^(2023) + (1/x^(2023) ) = 2cos 2023θ = 2cos 2023(30°) = 2cos [(12(168) + 7)×30°] = 2cos [360(168) + (30×70)] = 2cos 210 = −2cos (210 − 180) = −2cos 30° = −(√3)

ORx=cosθ+isinθx+1x=2cosθ=3θ=cos−132=30°x2023+1x2023=2cos2023θ=2cos2023(30°)=2cos[(12(168)+7)×30°]=2cos[360(168)+(30×70)]=2cos210=−2cos(210−180)=−2cos30°=−3

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