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- Feb 14, 2012

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

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

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

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- Thread starter
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- #1

- Feb 14, 2012

- 3,690

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

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

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

- Mar 31, 2013

- 1,309

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

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

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

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- Feb 14, 2012

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Hi

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

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!

- Mar 31, 2013

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Hello anemoneHikaliprasad,

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!

Thanks for the encouragement.

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- Feb 14, 2012

- 3,690

HeyHello anemone

Thanks for the encouragement.

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!

- Feb 15, 2012

- 1,967

Here is another solution:

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

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

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 $a,b,c$)

- Mar 31, 2013

- 1,309

neat and elegantHere is another solution:

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

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

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 $a,b,c$)

- Feb 15, 2012

- 1,967

Why, thank you!neat and elegant

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).