3. Simplify: \( \begin{array}{ll}\text { a. }\left(2 a^{2} b^{3} c\right)^{3} & \text { b. } 5 a^{4}-\left(2 a^{2}\right)^{2} \\ \text { 4. Simplify as far as possible: } \\ \left.\text { 4. } 4 c^{5}\right)^{2} \\ \text { a. } 3^{-1}-4^{-1} & \text { b. } 3^{-2}+3 \frac{1}{3}-\left(\frac{3}{10}\right)^{-1}\end{array} \)

Let's dive into the simplifications one step at a time! 1. For \((2 a^{2} b^{3} c)^{3}\): When raising a product to a power, you multiply the exponents. Thus, you get: \[ (2^3)(a^{2 \cdot 3})(b^{3 \cdot 3})(c^{3}) = 8a^{6}b^{9}c^{3} \] 2. For \(5 a^{4} - (2 a^{2})^{2}\): First, simplify \((2 a^{2})^{2} = 4 a^{4}\). Then, you have: \[ 5 a^{4} - 4 a^{4} = (5 - 4)a^{4} = 1a^{4} = a^{4} \] 3. For \((4 c^{5})^{2}\): Following the exponentiation rule, we simplify: \[ 4^2 (c^{5})^{2} = 16 c^{10} \] 4. For \(3^{-1} - 4^{-1}\): This translates to: \[ \frac{1}{3} - \frac{1}{4} = \frac{4 - 3}{12} = \frac{1}{12} \] 5. For \(3^{-2} + 3 \frac{1}{3} - \left(\frac{3}{10}\right)^{-1}\): Simplifying each term gives: \[ \frac{1}{9} + 3 \cdot \frac{1}{3} - \frac{10}{3} = \frac{1}{9} + 1 - \frac{10}{3} \] Converting \(1\) into a fraction with a denominator of \(9\): \[ \frac{1}{9} + \frac{9}{9} - \frac{30}{9} = \frac{1 + 9 - 30}{9} = \frac{-20}{9} \] Thus, the results are: 1. \( 8a^{6}b^{9}c^{3} \) 2. \( a^{4} \) 3. \( 16c^{10} \) 4. \( \frac{1}{12} \) 5. \( \frac{-20}{9} \)