factorise fully \( 6 a^{2} p x-4 a p^{2} y-6 a^{2} p y+4 a p^{2} x \)

To factorise the expression \( 6 a^{2} p x - 4 a p^{2} y - 6 a^{2} p y + 4 a p^{2} x \), we can start by grouping the terms: 1. Group terms involving \(x\) and \(y\): \[ (6 a^{2} p x + 4 a p^{2} x) + (-4 a p^{2} y - 6 a^{2} p y) \] 2. Factor out common factors in each group: \[ x(6 a^{2} p + 4 a p^{2}) - y(4 a p^{2} + 6 a^{2} p) \] 3. Notice that \(4 a p^2\) and \(6 a^2 p\) are common in both terms. We can factor out this common expression: \[ = (x - y)(6 a^{2} p + 4 a p^{2}) \] 4. Further simplifying \(6 a^{2} p + 4 a p^{2}\): \[ = 2 a p(3 a + 2 p) \] 5. Therefore, the fully factorised form of the expression is: \[ 2 a p (3 a + 2 p)(x - y) \]