If a,b,c are distinct real numbers such that `a+(1)/(b)=b+(1)/(c )=c+(1)/(a)`, then evaluate abc.

Options

A. `pmsqrt(2)`
B. `sqrt(2)-1`
C. `sqrt(3)`
D. `pm1`

Answer

To solve the problem where \( a, b, c \) are distinct real numbers such that

\[ a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a}, \]

we will denote this common value as \( k \). Thus, we can write the following equations:

1. \( a + \frac{1}{b} = k \)
2. \( b + \frac{1}{c} = k \)
3. \( c + \frac{1}{a} = k \)

From each of these equations, we can express \( a, b, \) and \( c \) in terms of \( k \):

1. From the first equation, rearranging gives:
\[ a = k - \frac{1}{b} \quad \text{(1)} \]

2. From the second equation, rearranging gives:
\[ b = k - \frac{1}{c} \quad \text{(2)} \]

3. From the third equation, rearranging gives:
\[ c = k - \frac{1}{a} \quad \text{(3)} \]

Next, we will equate the first two equations to eliminate \( k \):

From (1) and (2):
\[ k - \frac{1}{b} = k - \frac{1}{c} \]
This simplifies to:
\[ \frac{1}{b} = \frac{1}{c} \implies b = c \quad \text{(not possible since \( b \) and \( c \) are distinct)} \]

Now, we will equate the second and third equations:

From (2) and (3):
\[ k - \frac{1}{c} = k - \frac{1}{a} \]
This simplifies to:
\[ \frac{1}{c} = \frac{1}{a} \implies c = a \quad \text{(not possible since \( a \) and \( c \) are distinct)} \]

Next, we will equate the first and third equations:

From (1) and (3):
\[ k - \frac{1}{b} = k - \frac{1}{a} \]
This simplifies to:
\[ \frac{1}{b} = \frac{1}{a} \implies b = a \quad \text{(not possible since \( a \) and \( b \) are distinct)} \]

Since all three pairs lead to contradictions, we will look for a different approach.

Let’s multiply the three equations together:

\[ \left(a - b\right) \left(b - c\right) \left(c - a\right) = \left(\frac{1}{b} - \frac{1}{c}\right) \left(\frac{1}{c} - \frac{1}{a}\right) \left(\frac{1}{a} - \frac{1}{b}\right) \]

The right-hand side can be simplified using the identity \( \frac{1}{x} - \frac{1}{y} = \frac{y - x}{xy} \):

Thus, we have:
\[ \left(a - b\right) \left(b - c\right) \left(c - a\right) = \frac{(c - b)(a - c)(b - a)}{abc} \]

Now, we can see that both sides are equal, leading us to:

\[ abc = 1 \]

Thus, the product \( abc \) evaluates to:

\[ \boxed{1} \]