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

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Discussion Overview

The discussion revolves around the computation of the expression \((a+b)(a+c)(b+c)\) for the roots \(a, b, c\) of the cubic equation \(x^3-7x^2-6x+5=0\). The scope includes mathematical reasoning and problem-solving approaches related to algebra.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • Participants express appreciation for each other's problem-solving skills and methods, particularly highlighting the effectiveness of various approaches to algebraic challenges.
  • Some participants mention the enjoyment and elegance of the problem posed, indicating a positive engagement with the mathematical content.
  • There are mentions of different solutions being presented, though specific methods or calculations are not detailed in the posts.

Areas of Agreement / Disagreement

The discussion appears to be supportive and collaborative, with participants agreeing on the enjoyment of the problem and the effectiveness of each other's methods. However, specific mathematical disagreements or unresolved points regarding the computation itself are not explicitly stated.

Contextual Notes

No specific mathematical steps or assumptions are detailed in the posts, leaving the computation process and any potential limitations or dependencies on definitions unresolved.

Who May Find This Useful

Readers interested in algebraic problem-solving, particularly those looking for collaborative approaches to mathematical challenges, may find this discussion beneficial.

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

Compute $(a+b)(a+c)(b+c)$.
 
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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).
 

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