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Lordelo
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Joined: 29 Oct 2009
Posts: 21
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 Posted: 16 Nov 2009 10:03:14 am Post subject: |
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if (2^(1/3) - 1)^(1/3)=a^(1/3) + b^(1/3) + c^(1/3)
what's the exact value of a+b+c ? |
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DarkerLine
ceci n'est pas une |
Super Elite (Last Title)
Joined: 04 Nov 2003
Posts: 8328
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| Posted: 16 Nov 2009 11:41:19 am Post subject: |
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No way to tell. Two easy solutions to check are (a,b,c) = ((21/3 - 1)/27, (21/3 - 1)/27, (21/3 - 1)/27) and (21/3-1,0,0), which add up to different things.
The generalized mean inequality tells you that the sum of (21/3-1)/9 you get in the first of those two cases is the lowest possible sum, and that's about all you can say. |
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bfr
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Joined: 13 Feb 2006
Posts: 108
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| Posted: 18 Nov 2009 06:14:50 pm Post subject: |
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| ((21/3 - 1)/27, (21/3 - 1)/27, (21/3 - 1)/27) works? I may be wrong, but I think 21/3-1 is the answer. |
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DarkerLine
ceci n'est pas une |
Super Elite (Last Title)
Joined: 04 Nov 2003
Posts: 8328
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| Posted: 18 Nov 2009 06:34:34 pm Post subject: |
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bfr wrote: ((21/3 - 1)/27, (21/3 - 1)/27, (21/3 - 1)/27) works?
It does. 1/271/3 = 1/3, so ((21/3 - 1)/27)1/3 = (21/3 - 1)1/3/3. You take three of those and get (21/3 - 1)1/3.
Last edited by Guest on 18 Nov 2009 06:35:26 pm; edited 1 time in total |
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bfr
Member
Joined: 13 Feb 2006
Posts: 108
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| Posted: 18 Nov 2009 06:43:29 pm Post subject: |
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Oops, my mistake. I accidentally was doing a^3 + b^3 + c^3  |
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Weregoose
Authentic INTJ
Super Elite (Last Title)
Joined: 25 Nov 2004
Posts: 3976
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| Posted: 18 Nov 2009 09:26:57 pm Post subject: |
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Solve the expression for the three variables and sum the results together:
[font="times new roman"](−a − b + 6 m q s + 3 m r + 3 m t − 3 n q − 3 n s − 3 q t − 3 r s + p) +
(−a − c + 6 m q u + 3 m r + 3 m v − 3 n q − 3 n u − 3 q v − 3 r u + p) +
(−b − c + 6 m s u + 3 m t + 3 m v − 3 n s − 3 n u − 3 s v − 3 t u + p)
Where:
[font="times new roman"]m = p1/3
n = p2/3
p = 21/3 − 1
q = a1/3
r = a2/3
s = b1/3
t = b2/3
u = c1/3
v = c2/3
Alternatively:
[font="times new roman"]−2 (a + b + c) + 3 (p − v (q + s) − t (q + u) − r (s + u)) + 6 (m (r + t + v + q s + q u + s u) − n (q + s + u))
Another approach uses the definitions:
[font="times new roman"]m = p1/3
n = p2/3
p = 21/3 − 1
q = a1/3
r = b1/3
s = c1/3
t = r + s
u = q + s
v = q + r
Which gives:
[font="times new roman"]p − t (3 n − t (3 m − t)) +
p − u (3 n − u (3 m − u)) +
p − v (3 n − v (3 m − v))
That works out remarkably in TI-Basic as:
[font="verdana"]³[font="arial"]√({A,B,C
Ans-sum(Ans)+[font="verdana"]³[font="arial"]√([font="verdana"]³[font="arial"]√(2)-1
sum(Ans[font="verdana"]³
I'm sure that somewhere, somehow, this will all just collapse back down to A+B+C, but it was fun while it lasted. 
Last edited by Guest on 01 Jul 2010 10:26:04 am; edited 1 time in total |
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mr. sir
Newbie
Joined: 21 Feb 2008
Posts: 41
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Weregoose
Authentic INTJ
Super Elite (Last Title)
Joined: 25 Nov 2004
Posts: 3976
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| Posted: 28 Nov 2009 04:34:00 pm Post subject: |
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http://www.artofproblemsolving.com/ wrote: Problem 37: Write ³√(³√(2)-1) in the form ³√a + ³√b + ³√c, where a, b, c are rational numbers. This is not the same thing as asking for a + b + c. The one given in this thread is an entirely different problem!
If you change the wording of an assignment, you're bound to change its actual interpretation.
Last edited by Guest on 28 Nov 2009 04:38:48 pm; edited 1 time in total |
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mr. sir
Newbie
Joined: 21 Feb 2008
Posts: 41
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| Posted: 29 Nov 2009 04:03:54 pm Post subject: |
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Maybe, but:
1. Your solution was just a garble of variables that didn't really mean anything to me, and the answer found on the AOPS forum is an answer that has some tangible numbers to work with.
2. The problems' wordings were so remarkably similar I assumed that they came from the same place and one was just paraphrased, losing something in the translation.
3. The person who answered the question did find a+b+c.
Last edited by Guest on 29 Nov 2009 04:05:34 pm; edited 1 time in total |
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Weregoose
Authentic INTJ
Super Elite (Last Title)
Joined: 25 Nov 2004
Posts: 3976
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| Posted: 29 Nov 2009 04:39:03 pm Post subject: |
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Well, I knew in my post that the final answer would be a + b + c; I thought I'd have fun, anyhow.
But my last post wasn't directed at you. |
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