MHB How Do You Calculate (a+b)(a+c)(b+c) for Roots of a Cubic Equation?

  • Thread starter Thread starter anemone
  • Start date Start date
AI Thread Summary
To calculate (a+b)(a+c)(b+c) for the roots of the cubic equation x^3-7x^2-6x+5=0, the roots a, b, and c are utilized. Participants express appreciation for each other's problem-solving skills and methods. The discussion highlights the collaborative nature of solving mathematical challenges, with compliments exchanged among contributors. Acknowledgment of the original problem's creativity is also noted. The conversation emphasizes learning and sharing knowledge in mathematics.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Let $a, b, c$ be the roots of $x^3-7x^2-6x+5=0$.

Compute $(a+b)(a+c)(b+c)$.
 
Mathematics news on Phys.org
Re: Compute (a+b)(a+c)(b+c)

anemone said:
Let $a, b, c$ be the roots of $x^3-7x^2-6x+5=0$.

Compute $(a+b)(a+c)(b+c)$.

F(x) = x^3- 7x^2 – 6x + 5
now a+ b+c = 7 so a +b = 7-c, b+c = 7-a, a + c = 7- b
so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a)
again as a, b,c are roots
f(x) = (x-a)(x-b)(x-c)
so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a) = f(7) = 7^3 – 7 * 7^2 – 6*7 + 5 = - 37
 
Re: Compute (a+b)(a+c)(b+c)

kaliprasad said:
F(x) = x^3- 7x^2 – 6x + 5
now a+ b+c = 7 so a +b = 7-c, b+c = 7-a, a + c = 7- b
so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a)
again as a, b,c are roots
f(x) = (x-a)(x-b)(x-c)
so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a) = f(7) = 7^3 – 7 * 7^2 – 6*7 + 5 = - 37

Hi kaliprasad,

Thanks for participating and well done, kali! It seems to me you're quite capable and always have a few tricks up to your sleeve when it comes to solving most of my challenge problems!
 
Re: Compute (a+b)(a+c)(b+c)

anemone said:
Hi kaliprasad,

Thanks for participating and well done, kali! It seems to me you're quite capable and always have a few tricks up to your sleeve when it comes to solving most of my challenge problems!

Hello anemone

Thanks for the encouragement.
 
Re: Compute (a+b)(a+c)(b+c)

kaliprasad said:
Hello anemone

Thanks for the encouragement.

Hey kaliprasad,

I've been told that a compliment, written or spoken, can go a long way...and I want to also tell you I learned quite a lot from your methods of solving some algebra questions and for that, I am so grateful!
 
Re: Compute (a+b)(a+c)(b+c)

Here is another solution:

$(a+b)(a+c)(a+b) = a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 2abc$

$= a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 3abc - abc$

$= (a + b + c)(ab + ac + bc) - abc$

Now, $x^3 - 7x^2 - 6x + 5 = (x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc$

From which we conclude that:

$a + b + c = 7$
$ab + ac + bc = -6$
$abc = -5$

and so: $(a+b)(a+c)(a+b) = (7)(-6) - (-5) = -42 + 5 = -37$

(this solution is motivated by consideration of symmetric polynomials in $a,b,c$)
 
Re: Compute (a+b)(a+c)(b+c)

Deveno said:
Here is another solution:

$(a+b)(a+c)(a+b) = a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 2abc$

$= a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 3abc - abc$

$= (a + b + c)(ab + ac + bc) - abc$

Now, $x^3 - 7x^2 - 6x + 5 = (x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc$

From which we conclude that:

$a + b + c = 7$
$ab + ac + bc = -6$
$abc = -5$

and so: $(a+b)(a+c)(a+b) = (7)(-6) - (-5) = -42 + 5 = -37$

(this solution is motivated by consideration of symmetric polynomials in $a,b,c$)

neat and elegant
 
Re: Compute (a+b)(a+c)(b+c)

kaliprasad said:
neat and elegant

Why, thank you!

Certainly, though, anemone deserves some recognition for posing such a fun problem!

(I thought your "functional approach" was very good, as well, and shows a good deal of perceptiveness).
 
Back
Top