Question 1036823
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if (x+1/x)^2 = 3 then find the value of x^206 + x^200 + x^90 + x^84 + x^18 + x^12 + x^6 + 1  
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<U>Answer</U>. If {{{(x+1/x)^2}}} = {{{3}}} then the value of {{{x^206 + x^200 + x^90 + x^84 + x^18 + x^12 + x^6 + 1}}} is zero.


<U>Solution</U>

<pre>
If  {{{(x+1/x)^2}}} = {{{3}}}  then

{{{x^2}}} + {{{2}}} + {{{1/x^2}}} = {{{3}}},

{{{x^2}}} + {{{1/x^2}}} = {{{1}}},

{{{x^2 -1 + 1/x^2}}} = {{{0}}},   ( <--- Introduce new variable u = {{{x^2}}} )  ---> 

{{{u - 1 + 1/u}}} = {{{0}}},   ( <---multiply both sides by u)  ---> 

{{{u^2 -u + 1}}} = {{{0}}},    ( <--- solve by using the quadratic formula )  ---> 

{{{u[1,2]}}} = {{{1/2 +- i*(sqrt(3)/2)}}} = {{{cos(pi/3) +- i*sin(pi/3))}}},   --->

{{{x[1,2]}}} = {{{cos(pi/6) +- i*sin(pi/6)}}}.   ( since x = {{{sqrt(u)}}} )

In other words, if  {{{(x+1/x)^2}}} = {{{3}}}  then x is the primitive root of the degree 12 of 1.

Then

{{{x^90}}} = {{{-x^84}}},  so  {{{x^90 + x^84}}} = {{{0}}};

{{{x^18}}} = {{{-x^12}}},  so  {{{x^18 + x^12}}} = {{{0}}};

{{{x^6}}} = {{{-1}}},  so  {{{x^6 + 1}}} = {{{0}}},  and also

{{{x^206}}} = {{{-x^200}}},  so  {{{x^206 + x^200}}} = {{{0}}}.

Hence,  the value of {{{x^206 + x^200 + x^90 + x^84 + x^18 + x^12 + x^6 + 1}}} is zero.
</pre>Solved.