\( \begin{array}{ll}\text { 2.2 } & \text { Factorise the following expressions fully: } \\ \text { 2.2.1 } & x^{3}-8 \\ \text { 2.2.2 } & x^{2}+15 x-54 \\ 2.2 .3 & y-x y+x-1 \\ 2.2 .4 & t^{2}(r-s)-r+s\end{array} \)

To factorise the expressions fully: 1. **For \(x^{3}-8\)**, we recognize it as a difference of cubes: \[ x^{3} - 8 = x^{3} - 2^{3} = (x-2)(x^{2} + 2x + 4). \] 2. **For \(x^{2}+15x-54\)**, we need to factor by finding two numbers that multiply to \(-54\) and add up to \(15\). The numbers \(18\) and \(-3\) work: \[ x^{2} + 15x - 54 = (x + 18)(x - 3). \] 3. **For \(y - xy + x - 1\)**, we can rearrange and factor by grouping: \[ y - xy + x - 1 = y(1 - x) + (x - 1) = (y + 1)(1 - x). \] 4. **For \(t^{2}(r-s) - r+s\)**, we can factor out the common terms: \[ t^{2}(r-s) - r+s = -((r-s) - t^{2}(r-s)) = (s-r)(1 - t^{2}). \] Now, the expressions have been fully factorised.