QUESTION 3 In this section the laws which were investigated in cases 1 and 2 , will be applied. However, the application will be on different variables. There is no specific order in which laws are applied. Leave answers with positive exponents. NB: All your previous knowledge of integers, whole numbers, algebraic expressions, etc. must be implemented in this section. Simplify the exponential expressions below, WITHOUT using a calculator: \( 3.1\left(5 x y^{2}\right)^{2} \) \( \qquad \) \( \qquad \) \[ 3.2\left(2 a^{3} b\right)^{3} \times 3\left(4 a b^{4}\right)^{2} \] \( \qquad \) \( \qquad \) \( \qquad \) PI

When simplifying \( 3.1\left(5 x y^{2}\right)^{2} \), we start by squaring the expression inside the parentheses: \[ (5xy^{2})^{2} = 5^{2} \cdot x^{2} \cdot (y^{2})^{2} = 25x^{2}y^{4}. \] Now, multiply by 3.1: \[ 3.1 \cdot 25x^{2}y^{4} = 77.5x^{2}y^{4}. \] For the second expression \( 3.2\left(2 a^{3} b\right)^{3} \times 3\left(4 a b^{4}\right)^{2} \), we simplify each part: First, we calculate \( (2a^{3}b)^{3} = 2^{3} \cdot (a^{3})^{3} \cdot b^{3} = 8a^{9}b^{3} \), then multiply by 3.2: \[ 3.2 \cdot 8a^{9}b^{3} = 25.6a^{9}b^{3}. \] Next, for \( (4ab^{4})^{2} = 4^{2} \cdot a^{2} \cdot (b^{4})^{2} = 16a^{2}b^{8} \), and multiply by 3: \[ 3 \cdot 16a^{2}b^{8} = 48a^{2}b^{8}. \] Now, putting it all together: \[ 25.6a^{9}b^{3} \times 48a^{2}b^{8} = 25.6 \cdot 48 \cdot a^{11} \cdot b^{11}. \] Calculating \( 25.6 \cdot 48 \) gives \( 1228.8 \): Thus, the final result is \( 1228.8a^{11}b^{11} \).