1.4 Factorise the following: 1.4.1 \( \quad 8 x^{3}+1 \) 1.4.2 \( \quad 2 x^{2}-x-6 \) 1.4.3 \( \quad a^{2}-2 a b-2 b^{2}+a b \)

**1.4.1 Factorise \(8x^3+1\):** Recognise that \[ 8x^3 = (2x)^3 \quad \text{and} \quad 1 = 1^3. \] We use the sum of cubes formula: \[ a^3+b^3=(a+b)(a^2-ab+b^2). \] Let \( a=2x \) and \( b=1 \). Then: \[ 8x^3+1 = (2x+1)\left((2x)^2-2x\cdot1+1^2\right). \] Simplify the quadratic factor: \[ (2x)^2 = 4x^2,\quad 2x\cdot1 = 2x,\quad \text{and} \quad 1^2 = 1. \] Thus, \[ 8x^3+1 = (2x+1)(4x^2-2x+1). \] --- **1.4.2 Factorise \(2x^2-x-6\):** Multiply the coefficient of \( x^2 \) and the constant term: \[ 2 \times (-6) = -12. \] We need two numbers that multiply to \(-12\) and add up to \(-1\) (the coefficient of \( x \)). These numbers are \(-4\) and \(3\) because \[ -4+3=-1 \quad \text{and} \quad -4\times 3=-12. \] Rewrite the middle term: \[ 2x^2-x-6 = 2x^2-4x+3x-6. \] Now, group the terms: \[ (2x^2-4x)+(3x-6). \] Factor each group: \[ 2x^2-4x=2x(x-2),\quad 3x-6=3(x-2). \] Since both groups have a common factor \((x-2)\), factor it out: \[ 2x(x-2)+3(x-2)=(x-2)(2x+3). \] Thus, \[ 2x^2-x-6 = (2x+3)(x-2). \] --- **1.4.3 Factorise \(a^2-2ab-2b^2+ab\):** First, combine like terms: \[ a^2-2ab+ab-2b^2 = a^2-ab-2b^2. \] Think of this as a quadratic in \(a\): \[ a^2-ab-2b^2. \] We look for two numbers that multiply to \(-2b^2\) and add to \(-b\). Factorise by inspection: \[ a^2-ab-2b^2=(a-2b)(a+b). \] Verification by expanding: \[ (a-2b)(a+b)=a^2+ab-2ab-2b^2 = a^2-ab-2b^2. \] Thus, \[ a^2-2ab-2b^2+ab = (a-2b)(a+b). \]