Question

please answer this question ​

Answer 1

answer is 3(x-2y)(2y-3z)(3z-x)

Answer 2

Answer:

We know that,

a

3

+b

3

+c

3

=(a+b+c)(a

2

+b

2

+c

2

−ab−bc−ca)+3abc

If a+b+c=0, then a

3

+b

3

+c

3

=3abc

Now, given (x−2y)

3

+(2y−3z)

3

+(3z−x)

3

Here, a=(x−2y), b=(2y−3z), c=(3z−x) and a+b+c=(x−2y)+(2y−3z)+(3z−x)=0

Therefore

(x−2y)

3

+(2y−3z)

3

+(3z−x)

3

=3(x−2y)(2y−3z)(3z−x)