6. Simplify and write the following with positive indices. \( \begin{array}{lll}\text { a) } a^{-2} b^{4} & \text { b) }(2 x)^{-3} & \text { c) } 3 a^{2} b \times 6 a^{-5} b^{4}\end{array} \)

**a)** We start with \[ a^{-2}b^4. \] A negative exponent means taking the reciprocal. Thus, we rewrite \(a^{-2}\) as \[ \frac{1}{a^2}. \] So, the expression becomes \[ \frac{b^4}{a^2}. \] --- **b)** We have \[ (2x)^{-3}. \] A negative exponent indicates taking the reciprocal of the base raised to the positive exponent. Thus, \[ (2x)^{-3} = \frac{1}{(2x)^3}. \] Now, compute \((2x)^3\): \[ (2x)^3 = 2^3 \times x^3 = 8x^3. \] Therefore, \[ (2x)^{-3} = \frac{1}{8x^3}. \] --- **c)** Consider the product \[ 3a^2b \times 6a^{-5}b^4. \] First, multiply the numerical coefficients: \[ 3 \times 6 = 18. \] Next, combine the powers of \(a\): \[ a^2 \times a^{-5} = a^{2 + (-5)} = a^{-3}. \] Then, combine the powers of \(b\): \[ b \times b^4 = b^{1+4} = b^5. \] Thus, the product is \[ 18a^{-3}b^5. \] To express with positive indices, rewrite \(a^{-3}\) as \[ \frac{1}{a^3}. \] So, the final expression is \[ \frac{18b^5}{a^3}. \]