$\left[ \frac{1}{(\log_a bc) + 1} + \frac{1}{(\log_b ca) + 1} + \frac{1}{(\log_c ab) + 1} \right]$ is equal to

### Concept & Formula This problem is solved by rewriting the constant $1$ as a logarithm with a matching base to combine terms, and then utilizing the logarithmic reciprocal rule. $$ \log_x x = 1 $$ $$ \frac{1}{\log_x y} = \log_y x $$ ### Step-by-Step Solution * **Given:** We need to evaluate the sum of three algebraic fractions involving logarithms. $$ \frac{1}{\log_a bc + 1} + \frac{1}{\log_b ca + 1} + \frac{1}{\log_c ab + 1} $$ * **Calculation:** To combine the terms in the denominators, replace the $1$ in each fraction with a logarithm that matches the base of the adjacent logarithm. For the first term, $1 = \log_a a$. For the second term, $1 = \log_b b$. For the third term, $1 = \log_c c$. $$ \frac{1}{\log_a bc + \log_a a} + \frac{1}{\log_b ca + \log_b b} + \frac{1}{\log_c ab + \log_c c} $$ * Apply the product rule ($\log x + \log y = \log(xy)$) to merge the denominators. $$ \frac{1}{\log_a(abc)} + \frac{1}{\log_b(abc)} + \frac{1}{\log_c(abc)} $$ * Now, use the reciprocal rule to move the logarithms to the numerators. This swaps the base and the argument. Notice how this perfectly aligns all three bases to $(abc)$. $$ \log_{abc} a + \log_{abc} b + \log_{abc} c $$ * Apply the product rule once more to combine this sum into a single logarithm. $$ \log_{abc}(a \times b \times c) $$ $$ \log_{abc}(abc) $$ * Apply the identity rule. Since the base and argument are identical, the result is $1$. $$ 1 $$ ### Exam Strategy & Shortcut When you encounter symmetric, cyclical expressions in algebra or logarithms (where terms rotate uniformly like $a \rightarrow b \rightarrow c$), the expressions almost always collapse to $0$, $1$, or a simple product. Recognizing that the denominators will all normalize to $\log(abc)$ allows you to mentally flip them to base $abc$ and add them up to $1$ in just a few seconds without writing out the intermediate steps. ### Common Pitfall Students frequently get stuck because they do not realize the constant $1$ can be flexibly rewritten as $\log_a a$, $\log_b b$, etc. Leaving the $1$ as a standard integer makes it impossible to combine the terms in the denominator. Always look for ways to unify formats. ### Final Answer **Therefore, the correct answer is 1.**