SOLUTION: What is the sum of the real solutions of the equation (x+1)(x^2 + 1)(x^3 + 1)=30x^3?

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Question 1143178: What is the sum of the real solutions of the equation (x+1)(x^2 + 1)(x^3 + 1)=30x^3?
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

+%28x%2B1%29%28x%5E2+%2B+1%29%28x%5E3+%2B+1%29=30x%5E3
+%28x%5E3+%2B+x%5E2+%2B+x+%2B+1%29%28x%5E3+%2B+1%29=30x%5E3
+x%5E6+%2B+x%5E5+%2B+x%5E4+%2B+2+x%5E3+%2B+x%5E2+%2B+x+%2B+1=30x%5E3
+x%5E6+%2B+x%5E5+%2B+x%5E4+%2B+2+x%5E3+%2B+x%5E2+%2B+x+%2B+1-30x%5E3=0
+x%5E6+%2B+x%5E5+%2B+x%5E4+-+28+x%5E3+%2B+x%5E2+%2B+x+%2B+1=0...factor
%28x%5E2+-+3+x+%2B+1%29+%28x%5E4+%2B+4+x%5E3+%2B+12+x%5E2+%2B+4+x+%2B+1%29=0

if %28x%5E2+-+3+x+%2B+1%29+=0....use quadratic formula to ind solutions

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+....a=1,b=-3,c=1

x+=+%28-%28-3%29+%2B-+sqrt%28+%28-3%29%5E2-4%2A1%2A1+%29%29%2F%282%2A1%29+

x+=+%283+%2B-+sqrt%28+9-4+%29%29%2F2+
x+=+%283+%2B-+sqrt%28+5+%29%29%2F2+
real solutions are:
x%5B1%5D+=+3%2F2+%2B+sqrt%285%29%2F2
x%5B2%5D+=+3%2F2+-+sqrt%285%29%2F2

if x%5E4+%2B+4+x%5E3+%2B+12+x%5E2+%2B+4+x+%2B+1=0 you will get only complex solutions; so, disregard it
the sum of the real solutions will be:

x%5B1%5D%2Bx%5B2%5D+=+3%2F2+%2B+sqrt%285%29%2F2%2B3%2F2+-+sqrt%285%29%2F2

x%5B1%5D%2Bx%5B2%5D+=+3%2F2+%2B3%2F2+
x%5B1%5D%2Bx%5B2%5D+=+6%2F2+
x%5B1%5D%2Bx%5B2%5D+=+3+-> your answer



Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.

            The solution by @MathLover1 is fine.

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


The original equation was reduced by @Mathlover1 to 


    x%5E6+%2B+x%5E5+%2B+x%5E4+-28x%5E3+%2B+x%5E2+%2B+x+%2B+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%5E6+%2B+x%5E5+%2B+x%5E4+-28x%5E3+%2B+x%5E2+%2B+x+%2B+1 = %28x%5E2+-3x+%2B+1%29%2A%28x%5E4+%2B+4x%5E3+%2B+12x%5E2+%2B+4x+%2B+1%29.


The factor 

    x%5E4+%2B+4x%5E3+%2B+12x%5E2+%2B+4x+%2B+1 = %28x%2B1%29%5E4+%2B+6x%5E2 

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


So, only real roots of the quadratic polynomial  x%5E2+-3x+%2B+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.    ANSWER