Question

Without finding the cubes, factorise:
(x - 2y)” + (2y - 3z)3 + (3z – x).
​

Answer 1

Answer:

We know,

a^3 + b^2 + c^3 - 3abc = (a + b + c) (a^2 + b^2 + c^2 - ab - bc - ca)

If,

a + b + c = 0

Then,

a^3 + b^3 + c^3 = 3abc

Here,

a = x - 2y

b = 2y - 3z

c = 3z - x

Thus,

(x - 2y) + (2y - 3z) + (3z - x) = 0

And then,

(x - 2y)^3 + (2y - 3z)^3 + (3z - x)^3 = 3 (x - 2y) (2y - 3z) (3z - x)

______________

Hope it helps...!!!

Answer 2

Answer:

this answer is 0.

Step-by-step explanation:

because

x -2y

0 + 2y - 3z

+x + 0 +3z

answer is 0