Without finding the cubes, factorise :
`(x-2y)^(3)+(2y-3z)^(3)+(3z-x)^(3)`

Explanation

To factorize the expression \( (x - 2y)^3 + (2y - 3z)^3 + (3z - x)^3 \) without finding the cubes, we can use the identity for the sum of cubes. The identity states that:

\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]

In our case, we can identify:
- \( a = x - 2y \)
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