Question

This problem is related to Vieta's formulas on symmetric polynomials.
The cubic equation x^3 - x^2 - 4x +3 =0 has real solutions a, b and c. What is the value of a+b, b+c and a+c.

Answer 1

The good news is that it is a cubic, and thus a maximum of three solutions as to where the function crosses the x-axis.
If you do a google search it will show you exactly where the zeros are located. None of them are in easy integer value places...
Another hint would be for you to look up Vieta's formulas : )
Again, the zeros are not nice integer values.

Answer 2

Had to look this up, since I have not done this in 60 years (yes, I am that old)
For a cubic, Ax^3 + Bx^2 + Cx + D = 0
and we have the roots a, b, and c
a+b+c = -B/A
ab + ac + bc = C/A
abc = -D/A

so for yours, x^3 - x^2 - 4x + 3 = 0
a+b+c = 1
ab + ac + bc = -4
abc = -3

Wolfram shows 3 real roots, and they do satisfy my 3 equations
www.wolframalpha.com/input/?i=solve+x%5E3+-+x%5E2+-+4x+%2B+3+%3D+0
So you could find (a+b) etc correct to 4 decimal places.

I played around with my 3 equations to find exact values for (a+b) etc, but got into an algebraic mess. Perhaps somebody else can see what I can't at this point.