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Find a, b, and c if a + b + c = 5; a2 + b2 + c2 = 29, and a3 + b3 + c3 = 83.


This is an interesting problem, and some clever manipulation may be required. 

EQ 1: a + b + c = 5
EQ 2: a2 + b2 + c2 = 29
EQ 3: a3 + b3 + c3 = 83

Let's start by observing that if we square the first equation, we get:

EQ 4: a2 + b2 + c2 + 2ab + 2bc + 2ac = 25.

Subtracting (2) from (4), and simplifying, gives:

EQ 5: ab + bc + ac = -2

Now let's (3), and rewrite it with a3 + b3 factored:

EQ 6: (a + b)(a2 - ab + b2) +c3 = 83.

Ideally, it would be nice if we could rewrite (6) in terms of c. We note the following:

(1) can be rewritten as: a + b = 5 - c

(5) can be rewritten as ab = - 2 - bc - ac = -2 - c(a + b) = -2 - c(5 - c) = c2 - 5c - 2

(2) can be rewritten as a2 + b2 = 29 - c2

Substituting these into (6) gives us:

EQ 7: (5 - c)(29 - c2 - c2 + 5c + 2) + c3 = 83

We simplify this equation as follows:

(5 - c)(-2c2 + 5c + 31) + c3 = 83

2c3 - 15c2 - 6c + 155 + c3 = 83
3c3 - 15c2 - 6c + 72 = 0
c3 - 5c2 - 2c + 24 = 0

This is a cubic equation, so we naturally expect three answers. And this makes perfect sense, in light of the problem; the equations are symmetric with respect to a, b, and c, so once we find three values for c, we know that any combinations of those three values make ordered triples (a, b, c) that work.

I don't see a quick solution by grouping, so I resort to the rational root theorem until I find a value that works; then I factor the resulting quadratic to obtain -2, 3, and 4, in any order.

Thanks for the problem; that was a fun one!

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