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# Understand the problem

If  $\alpha$ is a root of  $x^2$ – x +1 = 0 , then $\alpha^{2018}$  + $\alpha^{-2018}$  is
##### Source of the problem
Sample Questions (MMA) : 2019
Medium
##### Suggested Book
Abstract Algebra – Dummit and Foote

Do you really need a hint? Try it first!

We have to use algebraic expression and expansion e.g-  $( x + 1) ^{2}$ = $x^2$ + 2x + 1 ; $( x +1)^{3}$ = $x^{3}$ + 3$x^2$ +3x + 1 Can you think about this kind of case a bit more ?
Note that $x^3$+ 1 = ($x+1$)($x^{2} – x + 1$) So, $\alpha$ is a root of $x^{2} – x + 1$ we have   ($\alpha^3$ + 1) = ($\alpha+1$)($\alpha^{2}$ – $\alpha$ + 1 ) = ( $\alpha$ + 1 ) = 0 Can we think along this line ?
$\alpha^{3}$ = -1   ||ly $\beta^{3}$ = -1 [ when $\beta$ is another root ($x^{2} – x + 1$) ] So, ($x^{2} – x + 1$) = ( x – $\alpha$ )( x – $\beta$) = $x^{2}$ – ($\alpha + \beta$)x + $\alpha$$\beta$. Can you get the expression of $\beta$ in terms of $\alpha$ ? Is the cloud getting clear now ?
$\alpha$ + $\beta$ = 1 $\alpha$$\beta$ =  1  implying   $\beta$  =  1/ $\alpha$  So, $\alpha^{2018}$  + $\alpha^{-2018}$  = $\alpha^{2018}$  + $\beta^{2018}$  Note that : $\alpha^{2}$  +  $\beta^{2}$ =  $\alpha + \beta )^{2}$ -2$\alpha$$\beta$  = 1 – 2$\alpha$$\beta$  = 1 -2 = -1   $\alpha^{2018}$ +   $\beta^{2018}$  =  $\alpha^{3×672}$ . $\alpha^{2}$ + $\beta^{3×672}$ . $\beta^{2}$ =  ($\alpha^{3})^{672}$.$\alpha^{2}$ + ($\beta^{3})^{672}$.$\beta^{2}$ = ($-1^{672}$).$\alpha^{2}$ + ($-1^{672}$).$\beta^{2}$ =  $\alpha^{2}$  + $\beta^{2}$   = -1

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