+288 (a) Prove that if the roots of x3+ax2+bx+c=0 form an arithmetic sequence, then 2a3+27c=9ab.
(b) Prove that if 2a3+27c=9ab, then the roots of x3+ax2+bx+c=0 form an arithmetic squence.
-------Thanks! 
+23254 a) Let the first root be "r", then since the roots form an arithmetic sequence, the other two roots are "r + d" and "r + 2d".
The factors will be: (x - r)(x - (r + d))(x - (r + 2d)) or (x - r)(x - r - d)(x - r - 2d)
Expanding (x - r)(x - r - d)(x - r - 2d) = x3 -3rx2 - 3dx2 + 3r2x + 6drx + 2d2x - r3 - 3dr2 - 2d2r
which means that a = -3r - 3d b = 3r2 + 6dr + 2d2 c = -r3 - 3dr2 - 2d2r
2a3 = -54r3 - 162r2d - 162rd2 - 54d3
27c = -27r3 - 81dr2 - 54d2r
9ab = -81r3 - 162r2d - 54d2r - 81r2d - 162d2r - 54d3
So, 2a3 + 27c = 9ab
[It wasn't easy ...]
