Question 397 of 455MediumMCQ
If \[\alpha ,\beta \in C\] are the distinct roots of the equation
x2 - x + 1 = 0, then \[{\alpha ^{101}} + {\beta ^{107}}\] is equal to :
Solutions · 1
Official solution
Given equation,
x2 - x + 1 = 0
Roots of this equation
x = \[{{1 \pm \sqrt 3 i} \over 2}\]
\[\therefore\,\,\,\] \[ \propto \, = \,{{1 + \sqrt 3 \,i} \over 2}\]
and \[\beta = \,{{1 - \sqrt 3 \,i} \over 2}\]
We know;
\[\omega = {{ - 1 + \sqrt 3 \,i} \over 2} = - \left( {{{1 - \sqrt 3 \,i} \over 2}} \right) = - \beta \]
and \[{\omega ^2} = {{ - 1 - \sqrt 3 \,i} \over 2} = - \left( {{{1 + \sqrt 3 \,i} \over 2}} \right) = - \propto \]
\[\therefore\,\,\,\] \[ \propto \, = - {\omega ^2}\] and \[\beta \, = \, - \omega \]
\[\therefore\,\,\,\] \[{ \propto ^{101}} + {\beta ^{107}}\]
\[ = {\left( { - {\omega ^2}} \right)^{101}} + {\left( { - \omega } \right)^{107}}\]
\[ = {\left( { - 1} \right)^{101}}.{\left( {{\omega ^2}} \right)^{101}} + {\left( { - 1} \right)^{107}}.{\left( \omega \right)^{107}}\]
\[ = - 1.{\left( {{\omega ^2}} \right)^{101}} - {\omega ^{107}}\]
\[ = - \left( {{\omega ^{202}} + {\omega ^{107}}} \right)\]
\[ = - \left( {{\omega ^{3.67}}.\omega + {\omega ^{3.35}}.{\omega ^2}} \right)\]
\[ = - \left( {\omega + {\omega ^2}} \right)\,\,\,\] [ as \[\,\,\,\] \[{\omega ^{3n}} = 1\]]
\[ = - \left( { - 1} \right)\] \[\,\,\,\,\,\,\] [as \[\,\,\,\] \[1 + \omega + {\omega ^2} = 0\] ]
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