Question 4: Problem solving 4.1 Determine the HCF of \( x \) and \( x^{3} \) 4.2 Determine the LCM of \( y \) and \( 2 y^{2} \) [Total: Section A \( =26 \) poin SECTION B: INTEGERS Question 5: Operations with integers Calculate the following, without a calculator using prime factors: \( 5.1 \sqrt[3]{-5832} \) \( 5.2 \sqrt{128} \) Question 6: Operations with integers Calculate the following, without a calculator and show all your calculations: \( 6.13-4 \times 6+2 \div 2 \) \( 6.2 \sqrt[3]{\left(4^{2}-3(2+1)\right)^{2}+3 \times 5} \)

To determine the HCF of \( x \) and \( x^{3} \), we look for the greatest common factor in their prime factorization. Since \( x \) is a common factor, the HCF is \( x \). Now, when finding the LCM of \( y \) and \( 2y^{2} \), we identify the maximum power of each factor involved. The LCM here is \( 2y^{2} \) as it takes the highest power of \( y \) involved. When calculating \( \sqrt[3]{-5832} \), first, factor \( -5832 \) into its prime components: \( -1 \times 2^3 \times 3^6 \). Thus, \( \sqrt[3]{-5832} = -18 \), since \( 2^3 \) can be easily cubed, and \( 3^6 \) becomes \( 3^2 \). For \( \sqrt{128} \), we rewrite it as \( \sqrt{2^7} = \sqrt{(2^3 \cdot 2^4)} = 4\sqrt{8} \), resulting in \( 8\sqrt{2} \). To solve \( 6.1 \) step by step: First, do the multiplication: \( 4 \times 6 = 24 \). Then proceed with \( 6.13 - 24 + 2 \div 2 \) where \( 2 \div 2 = 1 \); thus, \( 6.13 - 24 + 1 = -17 \). For \( 6.2 \), start with \( 4^2 - 3(2 + 1) = 16 - 9 = 7 \). Hence, we cube \( 7^2 + 15 \). Therefore, \( \sqrt[3]{49 + 15} = \sqrt[3]{64} = 4 \).