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Topic: What does this expression equal? (Read 5488 times) |
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pcbouhid
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What does this expression equal?
« on: Nov 25th, 2005, 10:20am » |
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If a + b + c = 0, what does the following expression equal?
[(b-c)/a + (c-a)/b + (a-b)/c] * [a/(b-c) + b/(c-a) + c/(a-b)].
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Michael Dagg
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Re: What does this expression equal?
« Reply #1 on: Nov 25th, 2005, 10:31am » |
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Looks like it = 9
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Michael Dagg
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pcbouhid
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Re: What does this expression equal?
« Reply #2 on: Nov 25th, 2005, 10:58am » |
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I read in a comment of William Wu something like this: "I don´t want just an answer; Í want to learn how to think with you".
And I agree totally with him!
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Michael Dagg
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Re: What does this expression equal?
« Reply #3 on: Nov 25th, 2005, 11:13am » |
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Well, then think about how I arrived at 9 by simple inspection.
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| « Last Edit: Nov 25th, 2005, 11:31am by Michael Dagg » |
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JohanC
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Re: What does this expression equal?
« Reply #4 on: Nov 25th, 2005, 11:55am » |
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Is there another way than just substitute c with -a-b and then do a lengthy but straightforward simplification ?
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Michael Dagg
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Re: What does this expression equal?
« Reply #5 on: Nov 25th, 2005, 12:35pm » |
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Investigate manipulating the condition a + b + c = 0 to form fractional expressions instead of using substitution.
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Michael Dagg
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Eigenray
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Re: What does this expression equal?
« Reply #6 on: Nov 25th, 2005, 8:39pm » |
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Substituting c=-(a+b),
[(b-c)/a + (c-a)/b + (a-b)/c] = f(a,b)/[ab(a+b)],
for some polynomial f of total degree 3. But we don't need to compute it to know that it's divisible by (a-b), (a+2b), and (2a+b), because it becomes 0 upon setting a=b, b=c, or c=a, respectively. It follows that
[(b-c)/a + (c-a)/b + (a-b)/c] = C(a-b)(a+2b)(2a+b)/[ab(a+b)].
Similarly,
[a/(b-c) + b/(c-a) + c/(a-b)] = g(a,b)/[(a+2b)(2a+b)(a-b)],
for some g. But if a=0, this becomes b/c-c/b = 0, for b=-c; thus g(a,b) is divisible by a. Similarly, it follows g(a,b) is divisible by a*b*(a+b). So
[a/(b-c) + b/(c-a) + c/(a-b)] = C'ab(a+b)/[(a+2b)(2a+b)(a-b)].
It follows
[(b-c)/a + (c-a)/b + (a-b)/c] * [a/(b-c) + b/(c-a) + c/(a-b)] = CC'
is constant. To evaluate the constant, one may take a=1, b=e2pi i/3, c=b2; then (b-c)/a = (c-a)/b = (a-b)/c, so the result is 9.
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Eigenray
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Re: What does this expression equal?
« Reply #7 on: Nov 25th, 2005, 9:03pm » |
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Maybe this is more constructive: let
x=a/(b-c), y=b/(c-a), z=c/(a-b).
Then
x+y = (ac-a2+b2-bc)/[(b-c)(c-a)] = [c(a-b)-(a+b)(a-b)]/[(b-c)(c-a)] = 2c(a-b)/[(b-c)(c-a)],
(x+y)/z = 2(a-b)2/[(b-c)(c-a)].
Taking cyclic permutations, it follows
[x+y+z][1/x+1/y+1/z] = 3 + 2[(a-b)3+(b-c)3+(c-a)3]/[(a-b)(b-c)(c-a)].
The numerator is a polynomial in a,b,c, of degree 3, and vanishes when a=b, b=c, or c=a. So the above is constant. Or to see this directly, use
u3 + v3 + (-u-v)3 = 3uv(-u-v).
Of course one can also do
1/y+1/z = (c-a)/b + (a-b)/c = [c2-ac+ab-b2]/[bc] = [(b-c)(a - (b+c))]/[bc] = 2a(b-c)/[bc],
x(1/y + 1/z) = 2a2/[bc],
so
[x+y+z][1/x+1/y+1/z] = 3 + 2[a3+b3+c3]/[abc] = 9.
(Another way to see the last equality is to note that (a3+b3+c3) is a symmetric polynomial, so is a polynomial in the elementary symmetric polynomials. Since it's degree 3, we have
a3+b3+c3 = A(a+b+c)3 + B(a+b+c)(ab+bc+ca) + C*abc = C*abc
for some C.)
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| « Last Edit: Nov 25th, 2005, 9:24pm by Eigenray » |
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Aryabhatta
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Re: What does this expression equal?
« Reply #8 on: Nov 26th, 2005, 1:30am » |
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Isn't (b-c)/a = (c-a)/b ?
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