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# Algebra Question.

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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?

Jun 2, 2021

#1
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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        ........!!!!

Jun 2, 2021
edited by CPhill  Jun 2, 2021
#2
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Thanks, CPhill !!

xCorrosive  Jun 2, 2021
#3
+121078
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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 ...)

CPhill  Jun 2, 2021
#4
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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 $y$.

xCorrosive  Jun 2, 2021
#5
+121078
+2

I tried  using  that hint, xC,  but I didn't  get  anywhere......maybe  someone else on here can steer  us  straight   !!!!

Jun 2, 2021
#6
+176
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=asodaksdopaksdopaksdpoasdk o0asod0- a=sd

ignore this message.

xCorrosive  Jun 3, 2021
edited by xCorrosive  Jun 9, 2021
edited by xCorrosive  Jul 31, 2021