Question 1208097
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If x + 1/x = {{{sqrt(3)}}}, find the value of x^18 + x^12 + 1.
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<pre>
If  x + 1/x = {{{sqrt(3)}}}, then

    {{{x^2}}} + 1 = {{{sqrt(3)*x}}}

    {{{x^2}}} - {{{sqrt(3)*x}}} + 1 = 0.


It is the general form of a quadratic equation  ax^2 + bx + c = 0
with a= 1, b= {{{-sqrt(3)}}}, c= 1.


Apply the quadratic formula to find the roots

    {{{x[1,2]}}} = {{{(-b +- sqrt(b^2-4ac))/(2a)}}} = {{{(sqrt(3) +- sqrt((sqrt(3))^2-4*1*1))/(2*1)}}} = {{{(sqrt(3) +- sqrt(3-4))/2}}} = {{{(sqrt(3) +- sqrt(-1))/2}}} = {{{(sqrt(3) +- i)/2}}}.


So,  {{{x[1]}}} = {{{(sqrt(3)+i)/2}}} = {{{cis(pi/6)}}};

     {{{x[2]}}} = {{{(sqrt(3)-i)/2}}} = {{{cis(-pi/6)}}}.



Thus  {{{x[1]}}}  is a complex number with the modulus of 1 and the argument of  {{{pi/6}}}.

      {{{x[2]}}}  is a complex number with the modulus of 1 and the argument of  {{{-pi/6}}}.



For  x = {{{x[1]}}} = {{{cis(pi/6)}}},  we have  

     {{{x^18}}} + {{{x^12}}} + 1 = {{{cis(18*(pi/6))}}} + {{{cis(12*(pi/6))}}} + 1 = {{{cis(3pi)}}} + {{{cis(2pi)}}} + 1 = -1 + 1 + 1 = 1.



For  x = {{{x[2]}}} = {{{cis(-pi/6)}}},  we have  

     {{{x^18}}} + {{{x^12}}} + 1 = {{{cis(18*(-pi/6))}}} + {{{cis(12*(-pi/6))}}} + 1 = {{{cis(-3pi)}}} + {{{cis(-2pi)}}} + 1 = -1 + 1 + 1 = 1.


So, for both values of roots, we have  {{{x^18}}} + {{{x^12}}} + 1 = 1.


At this point, the solution is complete.


<U>ANSWER</U>.  If  x + 1/x = {{{sqrt(3)}}},  then  {{{x^18}}} + {{{x^12}}} + 1 = 1.
</pre>

Solved.