# Assuming that (x-y)^2 + (y-z)^2 + (z-x)^2 = xyz, prove that x^3 + y^3 + z^3 is exactly divisible by x+y+z+6I need a proof for this question...

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Given : (x-y)2 +(y-z)2 +(x-z)2 = xyz

Require to prove :

x3 +y3 +z3 is divisible by x+ y+ z+ 6

Let us take the given equation :

(x-y)2 +(y-z)2 +(x-z)2 = xyz

=> x2+y2 -2xy+y2+z2 -2yz+x2+z2-2xz = xyz [ Expansion of (a-b)2 ]

=> 2x2+2y2+2z2 -2xy-2yz -2xz = xyz

=> 2(x2+y2+z2-xy-yz-xz) = xyz

=> x2+y2+z2-xy-yz-xz = (xyz/2) ------(1)

Now let us take the expression x3+y3+z3 from require to prove :

x3 +y3 +z3-3xyz = (x+y+z)( x2+y2+z2-xy -yz –xz)

[using the formula a3+b3+c3_3abc= (a+b+c)( a2+b2+c2b-bc -ac)]

=> x3 +y3 +z3-3xyz = (x+y+z)(xyz/2)

[since from eq.(1) we got: x2+y2+z2-xy-yz-xz = xyz/2 ]

=> x3 +y3 +z3= (x+y+z)(xyz/2)+3xyz

=> x3 +y3 +z3 = (xyz/2)(x+y+z+6) [ taking (xyz/2) common]

Since (x+y+z+6) is one of the factor of x3 +y3 +z3

Hence:

** x3 +y3 +z3 is divisible by (x+y+z+6) <--**** Proved**

By data (x-y)^2 + (y-z)^2 + (z-x)^2 = xyz.

Expand the LHS and rewrite :

x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2 = xyz

=>

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

=>

(x^2+y^2+z^2-xy-yz-zx) = xyz/2....(1)

We use the identity a^2+b^3+c^3- 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) to solve this problem.

x^3+y^3+z^3 -3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx) is an identity.

=> x^3+y^3+z^3 = (x+y+z)**(x^2+y^2+z^2-xy-yz-zx)+**3xyz...(2).

We use from (1), (x^2+y^2+z^2-xy-yz-zx) = xyz/2 to rewrite (2):

x^3+y^3+z^3 = (x+y+z){xyz/2)}+3xyz

x^3+y^3+z^3 = {x+y+z)/2+3}(xyz)

x^3+y^3+z^3 = **{x+y+z+6}**(xyz)/6

Therefore **{x+y+z+6}** **divides x^3+y^3+z^3**.