If abc = 5, what is the value of (\({1 \over 1 \ + \ a \ + \ b^{-1} }\) + \({1 \over 1 \ + \ b \ + \ 5 c^{-1}}\) +\(\frac{1}{1+\frac{c}{5}+a^{-1}}\) \(\) )?
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Understanding the Problem: Evaluating a Sum of Fractions
We are asked to find the value of a specific sum of three fractions, given the condition that the product of three variables \(a\), \(b\), and \(c\) is equal to 5, i.e., \(abc = 5\). The sum is given by:
\[ \left(\frac{1}{1 \ + \ a \ + \ b^{-1} }\right) + \left(\frac{1}{1 \ + \ b \ + \ 5 c^{-1}}\right) +\left(\frac{1}{1+\frac{c}{5}+a^{-1}}\right) \]To solve this, we need to simplify each term in the sum using the given condition \(abc = 5\).
Simplifying Each Term Using \(abc = 5\)
The condition \(abc = 5\) is the key to simplifying the fractions. Since \(abc=5\), none of \(a\), \(b\), or \(c\) can be zero. This allows us to use reciprocals and rearrange the equation.
Simplifying the Second Term
Let's consider the second term:
\[ T_2 = \frac{1}{1 \ + \ b \ + \ 5 c^{-1}} \]We know that \(c^{-1} = \frac{1}{c}\). From the condition \(abc = 5\), we can divide by \(c\) (since \(c \neq 0\)) to get \(ab = \frac{5}{c}\). Therefore, \(5 c^{-1} = 5 \cdot \frac{1}{c} = \frac{5}{c} = ab\).
Substituting \(5 c^{-1} = ab\) into the second term, we get:
\[ T_2 = \frac{1}{1 \ + \ b \ + \ ab} \]This term has a denominator \(1 + b + ab\).
Simplifying the First Term
Now let's look at the first term:
\[ T_1 = \frac{1}{1 \ + \ a \ + \ b^{-1} } \]To make its denominator similar to that of \(T_2\), let's try multiplying the numerator and the denominator by \(b\). Since \(b \neq 0\), this is a valid operation.
\[ T_1 = \frac{1 \cdot b}{(1 \ + \ a \ + \ b^{-1} ) \cdot b} = \frac{b}{1 \cdot b \ + \ a \cdot b \ + \ b^{-1} \cdot b} \]Since \(b \cdot b^{-1} = b \cdot \frac{1}{b} = 1\), we have:
\[ T_1 = \frac{b}{b \ + \ ab \ + \ 1} = \frac{b}{1 \ + \ b \ + \ ab} \]The first term now also has the denominator \(1 + b + ab\).
Simplifying the Third Term
Next, let's simplify the third term:
\[ T_3 = \frac{1}{1+\frac{c}{5}+a^{-1}} \]From the condition \(abc = 5\), we can divide by \(ab\) (since \(a \neq 0\) and \(b \neq 0\)) to get \(\frac{c}{5} = \frac{1}{ab}\). Also, \(a^{-1} = \frac{1}{a}\).
Substitute these into the third term:
\[ T_3 = \frac{1}{1+\frac{1}{ab}+\frac{1}{a}} \]To simplify this complex fraction and get the denominator \(1 + b + ab\), let's multiply the numerator and the denominator by \(ab\). Since \(ab \neq 0\), this is valid.
\[ T_3 = \frac{1 \cdot ab}{\left(1+\frac{1}{ab}+\frac{1}{a}\right) \cdot ab} = \frac{ab}{1 \cdot ab \ + \ \frac{1}{ab} \cdot ab \ + \ \frac{1}{a} \cdot ab} \] \[ T_3 = \frac{ab}{ab \ + \ 1 \ + \ b} = \frac{ab}{1 \ + \ b \ + \ ab} \]The third term also simplifies to a fraction with the denominator \(1 + b + ab\).
Calculating the Total Sum
Now that all three terms have the common denominator \(1 + b + ab\), we can add them easily:
\[ \text{Sum} = T_1 + T_2 + T_3 \] \[ \text{Sum} = \frac{b}{1 \ + \ b \ + \ ab} + \frac{1}{1 \ + \ b \ + \ ab} + \frac{ab}{1 \ + \ b \ + \ ab} \]Adding the numerators over the common denominator:
\[ \text{Sum} = \frac{b \ + \ 1 \ + \ ab}{1 \ + \ b \ + \ ab} \]The numerator \(b + 1 + ab\) is identical to the denominator \(1 + b + ab\). Assuming the denominator is not zero (which is implied by the problem resulting in a single numerical value), the fraction simplifies to 1.
\[ \text{Sum} = 1 \]Conclusion on the Value of the Sum
Through algebraic manipulation and utilizing the condition \(abc=5\), each term in the sum was transformed into a fraction with the common denominator \(1 + b + ab\). Summing these fractions resulted in \(\frac{1 + b + ab}{1 + b + ab}\), which simplifies to 1.
Revision Table: Breakdown of Term Simplification
Below is a summary of how each part of the sum was simplified:
| Original Term | Key Substitution / Manipulation | Simplified Form |
|---|---|---|
| \( \frac{1}{1 + a + b^{-1} } \) | Multiply numerator & denominator by \(b\) | \( \frac{b}{1 + b + ab} \) |
| \( \frac{1}{1 + b + 5 c^{-1}} \) | Use \(5c^{-1} = ab\) (from \(abc=5\)) | \( \frac{1}{1 + b + ab} \) |
| \( \frac{1}{1+\frac{c}{5}+a^{-1}} \) | Use \( \frac{c}{5} = \frac{1}{ab} \) and multiply numerator & denominator by \(ab\) | \( \frac{ab}{1 + b + ab} \) |
Additional Information: Techniques for Algebraic Sums
Problems involving sums of fractions with linked variables often require transforming the terms to find a common denominator. The given condition (\(abc=5\)) provides the necessary link between the variables \(a\), \(b\), and \(c\). Recognizing relationships like \(b^{-1} = \frac{ac}{5}\), \(c^{-1} = \frac{ab}{5}\), \(a^{-1} = \frac{bc}{5}\), \(\frac{c}{5} = \frac{1}{ab}\), etc., derived from the condition is crucial. Multiplying the numerator and denominator by a suitable expression is a standard technique to achieve a desired form or a common denominator, as demonstrated in this solution.
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