Correct Answer: B) $1$ Let's analyze the given expression step-by-step. The expression is: $$E = \frac{1}{1 + \log_a (bc)} + \frac{1}{1 + \log_b (ca)} + \frac{1}{1 + \log_c (ab)}$$ Step 1: Simplify the denominators using the logarithm property $\log_x x = 1$. We can rewrite the number $1$ in each denominator as a logarithm with the same base as the adjacent term. For the first term's denominator: $$1 + \log_a (bc) = \log_a a + \log_a (bc)$$ Using the product rule for logarithms, $\log_x M + \log_x N = \log_x (MN)$: $$\log_a a + \log_a (bc) = \log_a (a \cdot bc) = \log_a (abc)$$ Similarly, for the second term's denominator: $$1 + \log_b (ca) = \log_b b + \log_b (ca) = \log_b (b \cdot ca) = \log_b (abc)$$ And for the third term's denominator: $$1 + \log_c (ab) = \log_c c + \log_c (ab) = \log_c (c \cdot ab) = \log_c (abc)$$ Step 2: Substitute the simplified denominators back into the expression. The expression $E$ now becomes: $$E = \frac{1}{\log_a (abc)} + \frac{1}{\log_b (abc)} + \frac{1}{\log_c (abc)}$$ Step 3: Apply the change of base formula for logarithms, $\frac{1}{\log_x y} = \log_y x$. Using this property for each term: $$\frac{1}{\log_a (abc)} = \log_{abc} a$$ $$\frac{1}{\log_b (abc)} = \log_{abc} b$$ $$\frac{1}{\log_c (abc)} = \log_{abc} c$$ So, the expression $E$ transforms to: $$E = \log_{abc} a + \log_{abc} b + \log_{abc} c$$ Step 4: Apply the product rule for logarithms again. $$\log_{abc} a + \log_{abc} b + \log_{abc} c = \log_{abc} (a \cdot b \cdot c)$$ Step 5: Simplify the final logarithm. Since the base and the argument are the same, $\log_x x = 1$: $$E = \log_{abc} (abc) = 1$$ Therefore, the value of the expression $E$ is $1$.