Question

write the solution on paper ​

Answer 1

Answer:

We know the corollary: if a+b+c=0 then a

3

+b

3

+c

3

=3abc

Using the above identity taking a=x−y, b=y−z and c=z−x, we have a+b+c=x−y+y−z+z−x=0 then the equation (x−y)

3

+(y−z)

3

+(z−x)

3

can be factorised as follows:

(x−y)

3

+(y−z)

3

+(z−x)

3

=3(x−y)(y−z)(z−x)

Hence, (x−y)

3

+(y−z)

3

+(z−x)

3

=3(x−y)(y−z)(z−x)

Answer 2

I hope it's helpful to you.