Simplify \(\frac{a^3 \times b^{-8} \times c^7}{a^{-4} \times b^{-6} \times…

Concept: For any nonzero base, dividing like bases subtracts exponents: \(\frac{x^m}{x^n} = x^{m-n}\), and a negative exponent means the reciprocal, \(x^{-n} = \frac{1}{x^n}\). To simplify a product/quotient of several bases raised to powers, apply this quotient rule separately to each base’s numerator and denominator exponents.

Application:

1. Group the numerator and denominator exponents by base: a has exponents 3 (numerator) and -4 (denominator); b has -8 (numerator) and -6 (denominator); c has 7 (numerator) and 8 (denominator).

2. For a, subtract the denominator exponent from the numerator exponent: \(a^{3-(-4)} = a^{7}\).

3. For b: \(b^{-8-(-6)} = b^{-2}\).

4. For c: \(c^{7-8} = c^{-1}\).

5. Combine the three results into a single product: \(a^{7} \times b^{-2} \times c^{-1} = \frac{a^{7}}{b^{2}c}\).

Cross-check: substitute a = 2, b = 3, c = 5. The original expression evaluates to \(\frac{2^3 \times 3^{-8} \times 5^7}{2^{-4} \times 3^{-6} \times 5^8} = \frac{128}{45}\), and the simplified form \(\frac{a^7}{b^2c}\) gives \(\frac{2^7}{3^2 \times 5} = \frac{128}{45}\) — the two match, confirming the simplification.