1. **Problem (a): Simplify** $4p^3q^5 \times 6p^2q$ - We multiply the coefficients: $4 \times 6 = 24$ - For the variables with the same base, add the exponents: $p^{3} \times p^{2} = p^{3+2} = p^{5}$ - Similarly for $q$: $q^{5} \times q^{1} = q^{5+1} = q^{6}$ - So, the simplified expression is $$24p^{5}q^{6}$$ 2. **Problem (b): Simplify** $(5x^{2}y^{4})^{3}$ - Use the power of a product rule: $(ab)^{n} = a^{n}b^{n}$ - Apply the exponent to each factor: - Coefficient: $5^{3} = 125$ - $x^{2}$ raised to 3: $x^{2 \times 3} = x^{6}$ - $y^{4}$ raised to 3: $y^{4 \times 3} = y^{12}$ - So, the simplified expression is $$125x^{6}y^{12}$$ 3. **Problem (c): Factorise** $9a^{2} - b^{2}$ - Recognize this as a difference of squares: $A^{2} - B^{2} = (A - B)(A + B)$ - Here, $A = 3a$ and $B = b$ - So, factorisation is $$(3a - b)(3a + b)$$ **Final answers:** (a) $24p^{5}q^{6}$ (b) $125x^{6}y^{12}$ (c) $(3a - b)(3a + b)$