Question

factorize-> 8x³-(x-y)³ - (y-z)³ - (z+x)³​

Answer 1

Step-by-step explanation:

$8 {x}^{3} - ( {x}^{3} - 3 {x}^{2} y + 3x {y}^{2} - {y}^{3} ) - ( {y}^{3 } - 3 {y}^{2} z + 3 {z}^{2} y - {z}^{3} ) - ( {z}^{3} + 3 {z}^{2} x + 3 {x}^{2}z + {x}^{3} )$

Answer 2

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)