\( \frac { 2 ^ { 3 b + 2 } \cdot 7 ^ { 3 b - 2 } } { 14 ^ { 3 b } - 14 ^ { 3 b - 1 } \cdot 2 } \)

Let's simplify the expression step by step. The expression given is: \[ \frac { 2 ^ { 3 b + 2 } \cdot 7 ^ { 3 b - 2 } } { 14 ^ { 3 b } - 14 ^ { 3 b - 1 } \cdot 2 } \] First, we can rewrite the denominator: \[ 14^{3b} = (2 \cdot 7)^{3b} = 2^{3b} \cdot 7^{3b} \] \[ 14^{3b-1} \cdot 2 = (2 \cdot 7)^{3b-1} \cdot 2 = 2^{3b-1} \cdot 7^{3b-1} \cdot 2 = 2^{3b} \cdot 7^{3b-1} \] Now substituting back gives us: \[ 14^{3b} - 14^{3b-1} \cdot 2 = 2^{3b} \cdot 7^{3b} - 2^{3b} \cdot 7^{3b-1} \] Factoring out \(2^{3b}\): \[ = 2^{3b} (7^{3b} - 7^{3b-1}) = 2^{3b} (7^{3b-1}(7 - 1)) = 2^{3b} \cdot 7^{3b-1} \cdot 6 \] Substituting this back into the original expression gives us: \[ \frac{2^{3b+2} \cdot 7^{3b-2}}{2^{3b} \cdot 7^{3b-1} \cdot 6} \] Now we can simplify this: The \(2^{3b}\) in the denominator cancels with the \(2^{3b+2}\) in the numerator: \[ = \frac{2^2 \cdot 7^{3b-2}}{7^{3b-1} \cdot 6} \] Now, the \(7^{3b-1}\) in the denominator simplifies with \(7^{3b-2}\): \[ = \frac{4 \cdot 7^{(3b-2)-(3b-1)}}{6} = \frac{4 \cdot 7^{-1}}{6} = \frac{4}{6 \cdot 7} = \frac{2}{3 \cdot 7} \] Finally, we obtain the simplified result: \[ \frac{2}{21} \]