If the sum of 3 non-zero distinct real numbers a, b, and c is 2, and the two sets {a,b,c} and {1/a,1/b,1/c} are the same, what is the value of a^2+b^2+c^2?

Note: Two sets are the same if there is a one-to-one correspondence between their elements. For example, the sets {1,2,3} and {3,2,1} are the same. Neither of them are the same as {1,2,1}.

assuming a<=b<=c

a = 1/c
b = 1/b
c = 1/a
a+b+c=2
We see that b=±1, so a+c=1 or 3

If b=1,
a+ 1/a = 1
a^2 - a + 1 = 0
2a = 1±√3 i
Nope

If b = -1,
a + 1/a = 3
a^2 - 3a + 1 = 0
a = (3±√5)/2
c = 2/(3±√5) = (3∓√5)/2

a^2+b^2+c^2 = (14+6√5)/4 + 1 + (14-6√5)/4 = 8

To find the value of a^2 + b^2 + c^2, we need to first find the values of a, b, and c.

We are given that the sum of the three numbers is 2. So, we can write the equation:

a + b + c = 2 ---(1)

We are also given that the sets {a, b, c} and {1/a, 1/b, 1/c} are the same. This means that there is a one-to-one correspondence between their elements. In other words, for each element in one set, there is a corresponding element in the other set.

So, let's assume that a corresponds to 1/a, b corresponds to 1/b, and c corresponds to 1/c. We can write the equations:

a = 1/a ---(2)
b = 1/b ---(3)
c = 1/c ---(4)

Solving equations (2), (3), and (4), we get:

a^2 = 1 => a = 1 or a = -1
b^2 = 1 => b = 1 or b = -1
c^2 = 1 => c = 1 or c = -1

Since a, b, and c are distinct real numbers, none of them can be equal. Therefore, we have two possibilities:

Case 1: a = 1, b = -1, c = -1 (since we cannot have two negative values)

Case 2: a = -1, b = 1, c = 1

Now, let's find the value of a^2 + b^2 + c^2 for each case:

Case 1: a = 1, b = -1, c = -1

a^2 + b^2 + c^2 = (1)^2 + (-1)^2 + (-1)^2 = 1 + 1 + 1 = 3

Case 2: a = -1, b = 1, c = 1

a^2 + b^2 + c^2 = (-1)^2 + (1)^2 + (1)^2 = 1 + 1 + 1 = 3

Therefore, the value of a^2 + b^2 + c^2 is 3 in both cases.