Question 1143178
.


            The solution by @MathLover1 is fine.


            I came to clarify some key points of the solution and simplify it, where it is possible.



<pre>
The original equation was reduced by @Mathlover1 to 


    {{{x^6 + x^5 + x^4 -28x^3 + x^2 + x + 1}}} = {{{0}}}.


I do not know simple way to factor it.

But there are online tools for it in the Internet.


See, for example, the site 
https://www.mathportal.org/calculators/polynomials-solvers/polynomials-expanding-calculator.php


If you apply it, you will get factorization 

    {{{x^6 + x^5 + x^4 -28x^3 + x^2 + x + 1}}} = {{{(x^2 -3x + 1)*(x^4 + 4x^3 + 12x^2 + 4x + 1)}}}.


The factor 

    {{{x^4 + 4x^3 + 12x^2 + 4x + 1}}} = {{{(x+1)^4 + 6x^2}}} 

is always positive and, therefore, has no real roots.


So, only real roots of the quadratic polynomial  {{{x^2 -3x + 1}}} are of interest.


Since we are asked to find the sum of these roots, there is NO NEED to find the roots explicitly and individually.


By applying the Vieta's theorem, you get that their sum is equal to 3.    <U>ANSWER</U>
</pre>