Question

factorise (x-y)whole cube+(y-z)whole cube+(z-x)whole cube

Answer 1

check this out ....u will understand

Answer 2

Answer: The factor of $(x-y)^3+(y-z)^3+(z-x)^3 \text { is } 3(x-y)(y-z)(z-x)$

Step-by-step explanation:

To find: $(x-y)^3+(y-z)^3+(z-x)^3$

As we know

$a^3+b^3+c^3= 3abc \text { when } a+b+c=0$

let a= x-y , b= y-z , c= z-x

a+b+c= x-y+y-z+z-x = 0

therefore

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

Hence, the factor of $(x-y)^3+(y-z)^3+(z-x)^3 \text { is } 3(x-y)(y-z)(z-x)$