Question

Finding the cubes, factorise (x – y)^3 + (y – z)^3 + (z – x)^3.

Answer 1

Answer:

${(x - y)}^{3} + (y - z) {}^{3} + (z - x) {}^{3} \\ = {x}^{3} - 3xy(x - y) - {y}^{3} + {y}^{3} \\ - 3yz(y - z) - {z}^{3} + {z}^{3} - 3zx(z - x) - {x}^{3} \\ = - 3xy(x - y) - 3yz(y - z) - 3zx(z - x) \\ = - 3 {x}^{2} y + 3x {y}^{2} - 3 {y}^{2} z + 3y {z}^{2} \\ - 3 {z}^{2} x + 3z {x}^{2} \\ =$