2015/075) find $x^{63} + x^{44} + x^{37} + x^{31} + x^{26} + x^9 +6$

what is the value of the above expression for $(x + \dfrac{1}{x})^2 = 3$

we have
$x^2 + 2 + \dfrac{1}{x^2}= 3$
or $x^4 -x^2 + 1=0$
or $(x^4-x^2+1) (x^2+1)= 0$ as $x^2+1$ is not zero
or $x^6+1=0$

so $x^6= - 1$ and $x^{12} = 1$

so   $x^{63} + x^{44} + x^{37}+ x^{31} + x^{26} + x^9 +6$
= $x^3 + x^8 + x + x^7 + x^2 + x^9 + 6 $ taking mod 12 of exponent
= $x^3(x^6+1) + x^2(x^6+1) + x ( 1 + x^6) + 6$
= 6 
This I have picked from https://in.answers.yahoo.com/question/index?qid=20150801013839AAEmdz0