Let us factorise the following algebraic expressions:

(2x – y)³ – (x + y)³ + (2y – x)³

Let us consider (2x – y) = a, (2y – x) = b

a + b = (2x – y) + (2y – x) = x + y

So we can say that

(2x – y) ³ – (x + y) ³ + (2y – x) ³ = a³ + b³ – (a + b) ³ …Equation (i)

Using the identity of (a + b) ³ we get

(a + b) ³ = a³ + b³ + 3ab(a + b)

⇒ - 3ab(a + b) = a³ + b³ - (a + b)³

Using the above identity in Equation (i)

(2x – y) ³ – (x + y) ³ + (2y – x) ³ = - 3ab(a + b)

⇒ (2x – y) ³ – (x + y) ³ + (2y – x) ³ = - 3×(2x – y)× (2y – x)× (x + y)

⇒ (2x – y) ³ – (x + y) ³ + (2y – x) ³ = 3×(2x – y)× (x - 2y)× (x + y)

3(2x – y) (x - 2y) (x + y)