$$ (48)^{-\frac{2}{7}} \times (16)^{-\frac{5}{7}} \times (3)^{-\frac{5}{7}} = $$ $x$

Concept & Formula The problem tests the properties of exponents, specifically combining bases with identical powers. $$ a^n \times b^n = (ab)^n $$ $$ a^m \times a^n = a^{m+n} $$ Step-by-Step Solution * Notice that the last two terms share the exact same fractional exponent: $$ -\frac{5}{7} $$. * Group these terms using the rule $$ a^n \times b^n = (ab)^n $$: $$ (16)^{-\frac{5}{7}} \times (3)^{-\frac{5}{7}} = (16 \times 3)^{-\frac{5}{7}} = (48)^{-\frac{5}{7}} $$ * Now, substitute this simplified combined term back into the original expression: $$ (48)^{-\frac{2}{7}} \times (48)^{-\frac{5}{7}} $$ * Since the bases are now identical (48), simply add the exponents: $$ -\frac{2}{7} + \left(-\frac{5}{7}\right) = -\frac{7}{7} = -1 $$ * The expression simplifies to $$ (48)^{-1} $$. * Apply the negative exponent rule $$ a^{-1} = \frac{1}{a} $$ to yield $$ \frac{1}{48} $$. Exam Strategy & Shortcut Always scan for relationships between numbers before blindly factoring into primes. Spotting that $$ 16 \times 3 = 48 $$ instantly collapses the complicated terms into a single base (48), turning a 2-minute prime factorization slog into a 10-second mental addition of fractions. Look for synergy in the bases when exponents match! Common Pitfall A common mistake is completely breaking down 48, 16, and 3 into absolute prime factors (powers of 2 and 3) right at the start. While this method mathematically works, it is heavily prone to sign and arithmetic errors with the numerous fractional exponents and wastes valuable exam time. Final Answer **Therefore, the correct answer is $$ \frac{1}{48} $$.**