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Question Number 159189 by mathlove last updated on 14/Nov/21

	Answered by Rasheed.Sindhi last updated on 16/Nov/21
	$Let (A/B)=x x+(1/x)=1; x^(2019) +x^(−2019) =? ▶x+(1/x)=1 ⇒x^2 −x+1=0⇒(x+1)(x^2 −x+1)=0 ⇒x^3 +1=0⇒x^3 =−1 ∴ x is cuberoot of −1 ∴ x=−1,−ω,−ω^2 −1 is a root of x+1=0 −ω,−ω^2 are the roots of x^2 −x+1=0 where ω is cuberoot of unity. ▶x^(2019) +x^(−2019) =(−ω)^(2019) +(−ω)^(−2019) −(ω^(2019) +ω^(−2019) )=−({(ω)^3 }^(673) +{(ω)^3 }^(−673) ) =−(1^(673) +1^(673) )=−2 ((A/B))^(2019) +((B/A))^(2019) =−2$
	$L e t \frac{A}{B} = x$ $x + \frac{1}{x} = 1; x^{2019} + x^{- 2019} = ?$ $▸ x + \frac{1}{x} = 1$ $\Rightarrow x^{2} - x + 1 = 0 \Rightarrow (x + 1) (x^{2} - x + 1) = 0$ $\Rightarrow x^{3} + 1 = 0 \Rightarrow x^{3} = - 1$ $∴ x i s c u b e r o o t o f - 1$ $∴ x = - 1, - ω, - ω^{2}$ $- 1 i s a r o o t o f x + 1 = 0$ $- ω, - ω^{2} a r e t h e r o o t s o f x^{2} - x + 1 = 0$ $w h e r e ω i s c u b e r o o t o f u n i t y .$ $▸ x^{2019} + x^{- 2019} = {(- ω)}^{2019} + {(- ω)}^{- 2019}$ $- (ω^{2019} + ω^{- 2019}) = - ({{(ω)}^{3}}^{673} + {{(ω)}^{3}}^{- 673})$ $= - (1^{673} + 1^{673}) = - 2$ ${(\frac{A}{B})}^{2019} + {(\frac{B}{A})}^{2019} = - 2$
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