Mathematics 3 QUESTION 1 1.1 Factorise fully and simplify where possible Investigation 1.1.1 \( \quad x^{2}-y^{2} \) \( 1.1 .2 \quad 625 a^{4}-81 b^{8} \) 1.2 Write a general statement on how to factorise the expressions above QUESTION 2 2.1 Factorise fully and simplify where possible 2.1.1 \( \quad x^{2}+2 x y+y^{2} \) 2.1.2 \( \quad x^{2}-8 x+16 \) (2) 2.1.3 \( \quad 16 a^{4}-24 a^{2} b^{3}+9 b^{6} \) (2) 2.2 Write a general statement on how to factorise the expressions above (3) [10] QUESTION 3 3.1 Factorise fully and simplify where possible 3.1.1 \( \quad x^{3}+y^{3} \) 3.1.2 \( 8 a^{6}+729 \) 3.1.3 \( \quad(3 x+y)^{3}+(3 x-y)^{3} \)

### QUESTION 1 #### 1.1 Factorise fully and simplify where possible **1.1.1** We have \[ x^2 - y^2 \] This is a difference of two squares, so it can be factorised as \[ (x-y)(x+y). \] **1.1.2** We have \[ 625a^4-81b^8. \] Notice that \[ 625a^4=(25a^2)^2 \quad \text{and} \quad 81b^8=(9b^4)^2. \] Thus, the expression is a difference of two squares and can be factorised as \[ (25a^2-9b^4)(25a^2+9b^4). \] #### 1.2 General Statement Any expression of the form \[ A^2-B^2 \] can be factorised as \[ (A-B)(A+B). \] Both expressions above are examples of this rule. --- ### QUESTION 2 #### 2.1 Factorise fully and simplify where possible **2.1.1** We have \[ x^2+2xy+y^2. \] This is a perfect square trinomial, since \[ x^2+2xy+y^2=(x+y)^2. \] **2.1.2** We have \[ x^2-8x+16. \] Recognise that \[ x^2-8x+16=(x-4)^2. \] **2.1.3** We have \[ 16a^4-24a^2b^3+9b^6. \] Notice that \[ (4a^2-3b^3)^2 = 16a^4 - 2\cdot(4a^2)(3b^3) + 9b^6 = 16a^4-24a^2b^3+9b^6. \] Thus, \[ 16a^4-24a^2b^3+9b^6=(4a^2-3b^3)^2. \] #### 2.2 General Statement A perfect square trinomial of the form \[ A^2 \pm 2AB + B^2 \] can be factorised as \[ (A \pm B)^2. \] The expressions above are examples of this complete square factorisation. --- ### QUESTION 3 #### 3.1 Factorise fully and simplify where possible **3.1.1** We have \[ x^3+y^3. \] This is a sum of cubes, which factorises as \[ x^3+y^3=(x+y)(x^2-xy+y^2). \] **3.1.2** We have \[ 8a^6+729. \] Rewrite each term as a cube: \[ 8a^6 = (2a^2)^3 \quad \text{and} \quad 729 = 9^3. \] Using the sum of cubes formula \[ A^3+B^3=(A+B)(A^2-AB+B^2), \] with \(A=2a^2\) and \(B=9\), we obtain: \[ 8a^6+729=(2a^2+9)\Big((2a^2)^2-2a^2\cdot9+9^2\Big). \] Simplify the second factor: \[ (2a^2)^2=4a^4,\quad 2a^2\cdot9=18a^2,\quad 9^2=81. \] Thus, \[ 8a^6+729=(2a^2+9)(4a^4-18a^2+81). \] **3.1.3** We have \[ (3x+y)^3+(3x-y)^3. \] Let \[ A=3x+y \quad \text{and} \quad B=3x-y. \] Then by the sum of cubes formula: \[ A^3+B^3=(A+B)(A^2-AB+B^2). \] First, compute \(A+B\): \[ (3x+y)+(3x-y)=6x. \] Now, compute \(A^2-AB+B^2\). An alternative approach is to notice that the sum of cubes identity can be expressed as \[ A^3+B^3 = (A+B)^3-3AB(A+B). \] We already have \(A+B=6x\). Next, find \(AB\): \[ AB=(3x+y)(3x-y)=9x^2-y^2. \] Thus, \[ (3x+y)^3+(3x-y)^3 = (6x)^3-3(9x^2-y^2)(6x). \] Compute each term: \[ (6x)^3=216x^3, \] \[ 3(9x^2-y^2)(6x)=18x(9x^2-y^2). \] So, \[ 216x^3-18x(9x^2-y^2)=216x^3-162x^3+18xy^2=54x^3+18xy^2. \] Factor out \(18x\): \[ 54x^3+18xy^2=18x(3x^2+y^2). \] Thus, \[ (3x+y)^3+(3x-y)^3 = 18x(3x^2+y^2). \] --- ### Final Answers **QUESTION 1:** 1.1.1 \(\quad x^2-y^2=(x-y)(x+y)\) 1.1.2 \(\quad 625a^4-81b^8=(25a^2-9b^4)(25a^2+9b^4)\) 1.2 General Statement: For any \(A^2-B^2\), factorise as \((A-B)(A+B)\). **QUESTION 2:** 2.1.1 \(\quad x^2+2xy+y^2=(x+y)^2\) 2.1.2 \(\quad x^2-8x+16=(x-4)^2\) 2.1.3 \(\quad 16a^4-24a^2b^3+9b^6=(4a^2-3b^3)^2\) 2.2 General Statement: A perfect square trinomial \(A^2\pm2AB+B^2\) factorises as \((A\pm B)^2\). **QUESTION 3:** 3.1.1 \(\quad x^3+y^3=(x+y)(x^2-xy+y^2)\) 3.1.2 \(\quad 8a^6+729=(2a^2+9)(4a^4-18a^2+81)\) 3.1.3 \(\quad (3x+y)^3+(3x-y)^3=18x(3x^2+y^2)\)