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

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In summary, the conversation revolved around solving the problem of computing the product of the sums of the three roots of a cubic equation. The participants, kaliprasad and anemone, exchanged compliments and discussed different methods of solving the problem, ultimately arriving at the conclusion that both approaches were valid and admirable.
  • #1
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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  • #2
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
 
  • #3
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!
 
  • #4
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.
 
  • #5
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!
 
  • #6
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 $a,b,c$)
 
  • #7
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
 
  • #8
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).
 

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

What is the purpose of computing (a + b)(a + c)(b + c)?

The purpose of computing (a + b)(a + c)(b + c) is to simplify the expression and find the product of the three terms. This can be useful in solving algebraic equations and finding the area of certain geometric shapes.

What are the steps to compute (a + b)(a + c)(b + c)?

The steps to compute (a + b)(a + c)(b + c) are as follows: 1. Use the distributive property to expand the first two terms, resulting in a^2 + ac + ab + bc2. Use the distributive property again to expand the third term, resulting in a^2 + ac + ab + bc + b^2 + bc3. Combine like terms to simplify the expression, resulting in a^2 + b^2 + 2ab + 2ac + 2bc4. This simplified expression is the final answer for (a + b)(a + c)(b + c).

Can (a + b)(a + c)(b + c) be simplified further?

No, (a + b)(a + c)(b + c) cannot be simplified any further because all of the terms are already combined and there are no common factors that can be factored out.

What are some real-life applications of computing (a + b)(a + c)(b + c)?

Some real-life applications of computing (a + b)(a + c)(b + c) include finding the area of a rectangle with sides of length a and b, and the perimeter of a triangle with sides of length a, b, and c. This expression can also be used in solving algebraic equations and in financial calculations.

Can (a + b)(a + c)(b + c) be computed using different methods?

Yes, (a + b)(a + c)(b + c) can be computed using different methods such as the FOIL method, the distributive property, and the algebraic method. However, the end result will always be the same.

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