Factorise \( (x-1)^{4}-(2 x-5)^{2} \)

To factorise \( (x-1)^{4}-(2x-5)^{2} \), we can recognize this expression as a difference of squares. We have: \[ (a^2 - b^2) = (a-b)(a+b) \] Let \( a = (x-1)^{2} \) and \( b = (2x-5) \). The difference of squares can be rewritten as: \[ ((x-1)^{2} - (2x-5))((x-1)^{2} + (2x-5)) \] Now, we simplify each factor: 1. **First Factor:** \[ (x-1)^{2} - (2x-5) = (x^2 - 2x + 1) - (2x - 5) = x^2 - 4x + 6 \] 2. **Second Factor:** \[ (x-1)^{2} + (2x-5) = (x^2 - 2x + 1) + (2x - 5) = x^2 - 4 \] Now we can factor \( x^2 - 4 \) as it is also a difference of squares: \[ x^2 - 4 = (x-2)(x+2) \] Putting it all together, we have: \[ (x-1)^{4} - (2x-5)^{2} = (x^2 - 4x + 6)((x-2)(x+2)) \] Thus, the fully factorised form is: \[ (x^2 - 4x + 6)(x-2)(x+2) \]