Roots of a cubic polynomial

• Apr 3rd 2009, 02:59 AM
jacs
Roots of a cubic polynomial
Looking for a quick identity to deal with the sum of the cubes of the roots of a cubic.

I know that a² + b² + c² = (a + b + c)² - 2(ab + ac + bc)

and i am trying to figure out one for

a³ + b³ + c³

my teacher has it as a³ + b³ + c³ = (a + b + c)³ - 3(ab + ac + bc)

but when i tried expanding the RHS i got
a³ + b³ + c³ + 3a²b + 3a²c +3ab² + 3ac² + 3b²c + 3bc² - 3ab - 3ac - 3bc + 6abc
which isnt even close to the identity we are after

I have tried to come up it but haven't had any luck yet.

thanks for any help
• Apr 3rd 2009, 04:36 AM
Soroban
Hello, jacs!

Quote:

Looking for a quick identity to deal with the sum of the cubes of the roots of a cubic.

I know that: .$\displaystyle a^2 + b^2 + c^2 \;=\; (a + b + c)^2 - 2(ab + ac + bc)$

and i am trying to figure out one for: .$\displaystyle a^3 + b^3 + c^3$

My teacher has it as: .$\displaystyle a^3 + b^3 + c^3 \:=\:(a + b + c)^3 - 3(ab + ac + bc)$ . . . . Wrong!

$\displaystyle \text{We have: }\;(a+b+c)^3 \;=\;a^3+b^3+c^3+3a^2b+3ab^2+$ $\displaystyle 3b^2c+3bc^2+3a^2c+3ac^2+6abc$
. . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle \underbrace{\qquad\qquad\qquad\qquad\qquad\qquad\q quad\qquad\qquad\qquad}$
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle ^{\text{All of this is "extra"}}$

$\displaystyle \text{Therefore: }\;a^3+b^3+c^3 \;=\;(a+b+c)^3\; -$ $\displaystyle 3(a^2b+ab^2 +b^2c+bc^2+a^2c+ac^2+2abc)$

• Apr 3rd 2009, 04:37 AM
Mush
Quote:

Originally Posted by jacs
Looking for a quick identity to deal with the sum of the cubes of the roots of a cubic.

I know that a² + b² + c² = (a + b + c)² - 2(ab + ac + bc)

and i am trying to figure out one for

a³ + b³ + c³

my teacher has it as a³ + b³ + c³ = (a + b + c)³ - 3(ab + ac + bc)

but when i tried expanding the RHS i got
a³ + b³ + c³ + 3a²b + 3a²c +3ab² + 3ac² + 3b²c + 3bc² - 3ab - 3ac - 3bc + 6abc
which isnt even close to the identity we are after

I have tried to come up it but haven't had any luck yet.

thanks for any help

Why not write $\displaystyle a^3 + b^3 + c^3 = (a+b+c)^3 + f(a,b,c)$

Expand $\displaystyle (a + b + c)^3$ and see what is missing, and whatever is missing is what $\displaystyle f(a,b,c)$ is.

$\displaystyle (a+b+c)^3 = a^3 +b^3 + c^3+ 3a^2b + 3a^2c + 3ab^2 + 6abc + 3ac^2 + 3b^2c + 3bc^2$

Hence $\displaystyle f(a,b,c) = - 3a^2b - 3a^2c - 3ab^2 + 6abc - 3ac^2 - 3b^2c - 3bc^2$

$\displaystyle = -3(a^2b+a^2c +ab^2 + ac^2 + b^2c + bc^2 +3abc )$

and $\displaystyle a^3 + b^3 + c^3 = (a+b+c)^3 -3(a^2b+a^2c +ab^2 + ac^2 + b^2c + bc^2 +2abc )$
• Apr 3rd 2009, 04:51 AM
jacs
thanks to both of you for that, i have gotten that far, but what i forgot to add in my original post was that i need to break it down into terms of
a + b + c or
ab + ac + bc or
abc

as we are working with sum and products of roots of a cubic

from the looks of it though, there isnt going to be a nice easy fix
• Apr 3rd 2009, 12:32 PM
Mush
Quote:

Originally Posted by jacs
thanks to both of you for that, i have gotten that far, but what i forgot to add in my original post was that i need to break it down into terms of
a + b + c or
ab + ac + bc or
abc

as we are working with sum and products of roots of a cubic

from the looks of it though, there isnt going to be a nice easy fix

$\displaystyle a^3 + b^3 + c^3 = (a+b+c)^3 -3(a^2b+a^2c +ab^2 + ac^2 + b^2c + bc^2 +2abc )$

$\displaystyle = (a+b+c)^3 -3(a(ab+ac) + b(ab +bc) + c(ac+cb) +2abc)$