Question

without finding the cubes , factorise (x-2y)whole cube +(2y-3z) whole cube +(3z-x)whole cube​

Answer 1

Answer:

I hope it would help you

Answer 2

The factorization of $(x-2y)^3 +(2y-3z)^3 +(3z-x)^3$ is 3(x-2y)(2y-3z)(3z-x).

Explanation:

Identity in polynomials :

If $x+y+z=0$ , then $x^3+y^3+z^3=3xyz$   (1)

The given expression : $(x-2y)^3 +(2y-3z)^3 +(3z-x)^3$

Here , $(x-2y)+(2y-3z) +(3z-x)$

$=x-2y +2y-3z+3z-x=0$

Therefore by using (1), we get $(x-2y)^3 +(2y-3z)^3 +(3z-x)^3 = 3(x-2y)(2y-3z)(3z-x)$

Hence, the factorization of $(x-2y)^3 +(2y-3z)^3 +(3z-x)^3$ is 3(x-2y)(2y-3z)(3z-x).

# Learn more :

Find the cube of (x+2y+ 3z)

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