# 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

##### Topic

Quadratic Roots

##### Difficulty Level

Medium

##### Suggested Book

Abstract Algebra – Dummit and Foote

# Start with hints

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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it can be solved more easily.

Hi,

Yes, you are right but this method came into my mind first so I typed it down :). You can write your solution so that the readers can see that as well. Thanks a lot for seeing this. Keep reading and don’t forget to comment. Your valuable comments/appreciations inspire us to create more contents.

Regards,

Arnab