Question

prove that: (x-y)³+(y-z)³+(z-x)³=3(x-y)(y-z)(z-x)

Answer 1

Using the conditional identity

if a+b+c=0 then

${a}^{3} + {b}^{3} + {c}^{3} =3abc$

here a=x-y

b=y-z

c=z-x

a+b+c=(x-y) +(y-z) +(z-x)

x-y+y-z+z-x=0

so

${a}^{3} + {b}^{3} + {c}^{3} =3abc$

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

Answer 2

Answer:

hope it helps you mate !!!!!!