The function f(x)=x
3 - 2x
2 + 3x - 1 has distinct roots a, b, c.
Compute
f(a+b)f(b+c)f(c+a)
From the quadratic coefficient a+b+c=2. Then a+b=2-c, a+c=2-b, and b+c=2-a.
That motivates calculating g(x)=f(2-x). Then the three roots of g(x) will be a+b, b+c, and a+c.
g(x) = f(2-x)
= (2-x)^3 - 2*(2-x)^2 + 3*(2-x) - 1
= (8-12x+6x^2-x^3) - 2*(4-4x+x^2) + 3*(2-x) - 1
= 8 - 12x + 6x^2 - x^3 - 8 + 8x - 2x^2 + 6 - 3x - 1
= 5 - 7x + 4x^2 - x^3
The product of the roots of g(x) is the same as f(a+b)*f(b+c)*f(c+a). From the coefficients, the product of the roots of g(x) is -(5/(-1)) = 5.
Solution