Part 1:
Find any and all quadratic functions
f(x)=x
2+bx+c with roots {b,c}.
Part 2:
Find any and all pairs of quadratic functions
f1(x)=x2+b1x+c1 with roots {b2,c2} and
f2(x)=x2+b2x+c2 with roots {b1,c1}.
Part 3:
Find any and all trios of quadratic functions
f1(x)=x2+b1x+c1 with roots {b2,c2},
f2(x)=x2+b2x+c2 with roots {b3,c3}, and
f3(x)=x2+b3x+c3 with roots {b1,c1}.
The given conditions generate eight simultaneous equations leading to 8 SPECIFIC CASES .
It was impossible for me to solve these cases and arrive at a resolution.
However, comparing with Part-2 leads one to surmise that the initial sets of sextuplets would be:
(b1, b2, b3, c1, c2, c3)
= (-1/2, -1/2, -1/2, -1/2, -1/2, -1/2) or, (1,1,1,-2,-2,-2) or, (0,0,0,0,0,0)
Checking, the first set is invalidated out, so that:
(b1, b2, b3, c1, c2, c3) = (1,1,1, -2, -2, -2) and, (0,0,0,0,0,0) seems to be the required set of solutions.
**** I don't know about it, but there seems to exist some recurrence/recursion relation whereby the solutions to Part 3 could be derived from Part 1 and Part 2
Edited on June 16, 2022, 8:09 am
Puzzle Thoughts: Part 3
