Question 1142053
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            This problem has beautiful, nice, elegant and unexpected solution.



<pre>
Use the formula 


    {{{cos(3*theta)}}} = {{{4*cos^3(theta)}}} - {{{3*cos(theta)}}}.


This formula is valid for any angle {{{theta}}}.


    For its proof see my post 

        - <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.1142052.html>https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.1142052.html</A> 

    at this forum.


Let  {{{theta}}} = {{{2pi/9}}}  and let  x = {{{cos(2pi/9)}}}.


Notice that  {{{3*theta}}} = {{{(3*2*pi)/9}}} = {{{6pi/9}}} = {{{(2/3)*pi}}} = 120°.


Hence,  {{{cos(3*theta)}}} = {{{-1/2}}}.


From the other side,  {{{cos(3*theta)}}} = {{{4*cos^3(theta)}}} - {{{3*cos(theta)}}},  according to the formula above.


In other words, 


    {{{4x^3}}} - {{{3x}}} = {{{-1/2}}}.


Multiplying by 2 both sides and simplifying, you get


    {{{8x^3}}} - {{{6x}}} + {{{1}}} = 0.


It means that   x = {{{cos(2pi/9)}}}  is the solution of the given equation.
</pre>

The proof is completed.