Question

without finding the cubes simplify:- (x-y)³ +(y-z)³ +(z-x)³/ (x-y)² + (y-z)² + (z-x)²​

Answer 1

Answer:

(x-Y) +(y-z) +(z-x)

Step-by-step explanation:

We have,

(x-y)^3 + (y-z) ^3 +(z-x) ^3 - - - - - - 1

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

Equation 1 can be written as,

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

Similarly,

Equation 2 can be written as,

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

Then,

Dividing eq. 1/ eq. 2

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

We get,

(x-y) + (y-z) + (z-x)