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