Roots of a cubic polynomial

rock.freak667

1. The problem statement, all variables and given/known data
In the equation [tex]x^3+ax^2+bx+c=0[/tex]
the coefficients a,b and c are all real. It is given that all the roots are real and greater than 1.
(i) Prove that [tex]a<-3[/tex]
(ii)By considering the sum of the squares of the roots,prove that [tex]a^2>2b+3[/tex]
(iii)By considering the sum of the cubes of the roots,prove that [tex]a^3<-9b-3c-3[/tex]


2. Relevant equations

If the roots are A,B and C then A+B+C = a/1=a
ABC= -c/a
AB+AC+BC= b/a

3. The attempt at a solution

I do not know if there are any other formula for the squares/cubes of roots other than the ones i stated above; If there are any simpler ones please tell me.
I got out parts (ii) by taking (A+B+C)=a and appropriately squaring it, but I was unable to get out parts (i) and (iii), could someone please help me prove it..thanks

ZioX

Since the roots are real we know the polynomial factors into [itex](x-r_1)(x-r_2)(x-r_3)[/itex]. Look at how [itex](x-r_1)(x-r_2)(x-r_3)[/itex] multiplies out and look at a b and c in terms of the roots. For example, we know that c must be negative as [itex]-r_1r_2r_3=c<0[/itex]. We actually know [itex]c<-1[/itex] as each of these roots are greater than 1.

rock.freak667

thanks, I will re-try it and see now

dextercioby

i) is trivial since A+B+C=-a (see the minus sign, for more see http://en.wikipedia.org/wiki/Vieta_relations)

ii)is doable as you said.

dextercioby

The point iii) is really tricky.

[tex] (A+B+C)^3 = A^3 +B^3 +C^3 -3ABC +3(A+B+C)(AB+AC+BC) [/tex]

which means

[tex] -a^3 =A^3 +B^3 +C^3 +3c -3ab > 3+3c+9b [/tex] ,

where i used the fact that the sum of the cubes is larger than 3 and the fact that a is smaller than -3.

Multiply by -1 and you're done.