Question

without finding cubes, factorise:
1.(x-y)3 +(y-z)3 +(z-x)3
2.(x-2y)3 +(2y-3z)3 + (3z-x)3

Answer 1

when  a+b+c = 0  we know that  a³ + b³ + c³ =  3 a b c

(x-y)³ + (y-z)³ + (z - x)³
     = 3 (x - y) (y -z) (z- x)
     = 3 (xy - x z -y² + yz) (z -x)
     = 3 (xyz - x² y - x z² + x² z - y² z + x y² + y z² - xyz)
     = 3 [ x² (z - y) + z² (y - x)  + y² (x - z) ]
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similarly  x - 2y + 2y - 3z + 3z - x = 0

hence,  (x-2y)³ + (2y - 3z)³ + (3z-x)³ = 3 (x -2y) (2y -3z)(3z - x)
 

Answer 2

(x-y)^3 + (y-z)^3 + (z-x)^3  

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

= (z-x) [ (z-x)^2 - (x-y)^2 - (y-z)^2 + (x-y)(y-z) ]  

= (z-x) [ -2zx - 2y^2 + 2xy + 2yz + (x-y)(y-z) ]  

= (z-x) [ 2x(y-z) - 2y(y-z) + (x-y)(y-z) ]  

= (z-x) [ 2(x-y)(y-z) + (x-y)(y-z) ]  

= 3 (x-y)(y-z)(z-x)