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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? 

 

Thanks in advance!

 Jun 2, 2021
 #1
avatar+121056 
+2

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        ........!!!!

 

 

 

cool cool cool

 Jun 2, 2021
edited by CPhill  Jun 2, 2021
 #2
avatar+181 
+1

Thanks, CPhill !!

xCorrosive  Jun 2, 2021
 #3
avatar+121056 
+1

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 ...)

 

 

cool cool cool

CPhill  Jun 2, 2021
 #4
avatar+181 
+1

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
avatar+121056 
+2

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

 

 

cool cool cool

 Jun 2, 2021
 #6
avatar+181 
0

=asodaksdopaksdopaksdpoasdk o0asod0- a=sd

 

ignore this message. 

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

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