+179 1. Prove that if the roots of
$$ x^3+ax^2+bx+c = 0$$
form an arirthmetic sequence, than $2a^3+27c=9ab$.
2. Prove that if $2a^3+27c=9ab$, then the roots of
$$ x^3 + ax^2+bx+c = 0 $$
form an arithmetic sequence.
So far I've wrote the roots as $r$, $r+d$, and $r+2d$. Should I use Vieta's formulas? How should I continue?
Thanks in advance!
+129849 1. Let the roots = n - d, n , n + d
The sum of the roots = 3n = - a ⇒ a = -3n ⇒ a^3 = -27n^3 ⇒ 2a^3 = -54n^3
The product of the roots, taken 2 at a time are
n ( n - d) + n ( n +d) + (n + d) ( n - d) = b
3n^2 - d^2 = b
And the product of all three roots = - c
n (n -d)(n + d) = - c
n ( n^2 - d^2) = - c
n^3 - nd^2 = - c
nd^2 - n^3 = c
27nd^2 - 27n^3 = 27c
So
2a^3 + 27c = 9ab ???
-54n^3 + (27nd^2 - 27n^3) = 9 (-3n) (3n^2 - d^2)
-81n^3 + 27nd^2 = -81n^3 + 27nd^2 ........!!!!
+129849 THX.....I made an edit to correct a slight error, but it works out OK....
I'm not exactly sure how to prove the second one, but....if some idea hits me, I'll give it a try !!!! (Sorry ...)
+179 I have a hint that they provided.
Let $y = x + \frac{a}{3}$ and rewrite $x^3 + ax^2 + bx + c = 0$ in terms of $y$.
+129849 I tried using that hint, xC, but I didn't get anywhere......maybe someone else on here can steer us straight !!!!
