Seemingly Random Polynomial

We have, $(x+\frac{1}{x})^2=3$ $\Rightarrow (x+\frac{1}{x})=√3$ $\Rightarrow (x^3+\frac{1}{x^3})=0$

Further it can be seen that for every $n \in N$ , $(x^{3n}+\frac{1}{x^{3n}})=0 \space (Why)$

Now our required expression is, $x^{37}+ x^{31}+ x^{44}+ x^{26}+ x^{63}+ x^{9}+6$ $=x^{34}(x^3+\frac{1}{x^3})+ x^{35}(x^9+\frac{1}{x^9})+ x^{36}(x^{27}+\frac{1}{x^{27}})+6$ $= x^{34}(0)+ x^{35}(0)+ x^{36}(0)+6=\boxed{6}$

Any other solutions are welcomed for the sake of variety.

$(x^2+\frac{1}{x^2})^2=3$

or, $x^4-x^2+1=0$

or, $(x^2+1)(x^4-x^2+1)=0$

or, $x^6+1=0$ or, $x^6=-1$

So, $x^{63}+x^{44}+x^{37}+x^{31}+x^{26}+x^{9}+6$

$=x^{3}-x^{2}+x-x+x^{2}-x^{3}+6=\boxed{6}$

Recall that $\cos \theta = \frac{e^{i\theta}+e^{-i\theta}}{2}$ . Substituting in $x = e^{i\theta}$ gives

$\cos \theta = \frac{\sqrt{3}}{2}$

Since the ratio of the adjacent side to the hypotenuse in a 30-60-90 triangle from the angle $\pi/6$ is $\frac{\sqrt{3}}{2}$ , we can conclude

$x = e^{i\theta} = e^{i\pi/6} = (-1)^{1/6}$

It can be easily seen that $x^{6m+n} = (-1)^{m}x^{n}$ . Hence, our given expression is equal to

$x^{3}-x^{2}+x-x+x^{2}-x^{3}+6 = \boxed{6}$

$(x + \dfrac{1}{x})^2 = 3$

$=> x^2 + \dfrac{1}{x^2} = 1$

$= > x^n = x^{n-2} - x^{n-4}$ by multiplying $x^{n-2}$ on both sides

Hence, the above also yields $= > x^{n-2} = x^{n-4} - x^{n-6}$ , where substituting it back in to the previous equation yields

$x^n = -x^{n-6}$

Using this, the expression :

$x^{63} + x^{44} + x^{37} + x^{31} + x^{26} + x^{9} + 6$

$= x^{3} - x^{2} + x - x + x^2 - x^3 + 6$

$= \boxed{6}$

Obviously, From the problem, we have: $\left( x+\frac { 1 }{ x } \right) ^{ 2 }={ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2=3\Rightarrow { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } =1$

Therefore: ${ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } =\left( x+\frac { 1 }{ x } \right) \left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } -1 \right) =\left( x+\frac { 1 }{ x } \right) \cdot 0=0$ . Furthermore, we can prove by induction:

$\left( { x }^{ \left( { 3 }^{ n } \right) }+\frac { 1 }{ { x }^{ \left( { 3 }^{ n } \right) } } \right) =0$ , when $\left( x+\frac { 1 }{ x } \right) ^{ 2 }=3$ and $n\in N$

Now, our expression is: ${ x }^{ 63 }+{ x }^{ 44 }+{ x }^{ 37 }+{ x }^{ 31 }+{ x }^{ 26 }+{ x }^{ 9 }+6$ $={ x }^{ 35 }\left( { x }^{ 9 }+\frac { 1 }{ { x }^{ 9 } } \right) +{ x }^{ 36 }\left( { x }^{ 27 }+\frac { 1 }{ { x }^{ 27 } } \right) +{ x }^{ 34 }\left( { x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } \right) +6$ $=0+0+0+6$ $=6$

This will be easy if we put this on the complex plane.

$(x + \frac{1}{x})^2 = 3$

$x^2 + 2 + \frac{1}{x^2} = 3$

$x^4 - x^2 + 1 = 0$

The roots will be $cis(\frac{\Pi }{6}$ ), $cis(\frac{5\Pi}{6}$ ), $cis(\frac{7\Pi}{6}$ ), and $cis(\frac{11\Pi}{6}$ ).

Then substitute to the expression, (let's use $cis(\frac{\Pi }{6}$ ))

$x^{63} + x^{44} + x^{37} + x^{31} + x^{26} + x^9 + 6$

$= cis(63 \times \frac{\Pi }{6}) + cis(44 \times \frac{\Pi }{6}) + cis(37 \times \frac{\Pi }{6}) + cis(31 \times \frac{\Pi }{6}) + cis(26 \times \frac{\Pi }{6}) + cis(9 \times \frac{\Pi }{6}) + 6$

$= cis(\frac{21\Pi }{2}) + cis(\frac{22\Pi }{3}) + cis(\frac{37\Pi }{6}) + cis(\frac{31\Pi }{6}) + cis(\frac{13\Pi }{3}) + cis(\frac{3\Pi }{2}) + 6$

$= cis(\frac{1\Pi }{2}) + cis(\frac{-2\Pi }{3}) + cis(\frac{1\Pi }{6}) + cis(\frac{-1\Pi }{6}) + cis(\frac{1\Pi }{3}) + cis(\frac{-1\Pi }{2}) + 6$

$= [cos(\frac{1\Pi }{2}) + cos(\frac{-2\Pi }{3}) + cos(\frac{1\Pi }{6}) + cos(\frac{-1\Pi }{6}) + cos(\frac{1\Pi }{3}) + cos(\frac{-1\Pi }{2}) + 6] + \iota[sin(\frac{1\Pi }{2}) + sin(\frac{-2\Pi }{3}) + sin(\frac{1\Pi }{6}) + sin(\frac{-1\Pi }{6}) + sin(\frac{1\Pi }{3}) + sin(\frac{-1\Pi }{2})]$

$= 0$

Using the other roots will also lead to 0 .

$x^2+\dfrac 1 {x^2}=1.~~~~~~~~{\color{#3D99F6}{\implies~x^4=x^2-1.....(A)}}.~~~~~~~~\therefore~x^8=x^4-2x^2+1= - x^2.....(B).\\ \therefore~x^6= - 1......(C).~~~~~~~~~~~{\color{#3D99F6}{\implies~x^{12}=1....(D)}},~~~~~~~~~~~and~~~~~~~~x^3= i......(E)\\ {\color{#3D99F6}{By~(D)~and~(E)~~x^{63}=(x^{12})^5*x^3= i } }~~~~~~~~ By~(D)~and~(E)~~x^{44}=(x^{12})^3*x^8= - x^2 \\ By~(D)~~~x^{37}=(x^{12})^3*x= x~~~~~~~~~~~\color{#3D99F6}{By~(D)~and~(C)~~x^{31}=(x^{12})^2*x^6x= - x}\\ {\color{#3D99F6}{~(D)~~~x^{26}=(x^{12})^2*x^2= x^2}}~~~~~~~~~By~(C)~and~(E)~~x^9=x^{6}*x^3= - i\\ \therefore~ x^{63} + x^{44} + x^{37} + x^{31} + x^{26} + x^{9} + 6\\ =~~~i~~~~+ - x^2 ~~+~~x~~~ - ~x ~~~+ x^2~~- i~~+~~6=\color{#D61F06}{6}$

evaluate x^63+x^44+x^37+x^31+x^26+x^9+6, (x+1/x)^2=3 (wolframalpha substitution case)

x^63+x^44+x^37+x^31+x^26+x^9+6=6

Try to find roots take those comples roota in euler form....put it in equation

Solve to get x=iw,Substitute x=iw in eq to get 6 where w is cube root of unity.

Here we get for $\Large (x +1/x)^2 = 3$ we get $\Large {x^2}$ = - $\Large {\omega}$ [where $\omega$ is the cube root of unity ]

By further Solving $\large x^{63} + x^{44} + x^{37} + x^{31} + x^{26} + x^9 + 6$

And putting the values care fully we get all cut down except 6. Hence Ans = $6$

x^12=1, so just reduce the exponents.