if x^2+y^2+z^2=xy+yz+zx then find the value of x^3+y^3+z^3

Given-

x^2+y^2+z^2=xy+yz+zx ---------(1)
multiplying by (x+y+z) on both sides-
x^2+y^2+z^2 (x+y+z)=(xy+yz+zx ) (x+y+z)
expand the above equations-
x^3+y^3+z^3+xy^2+x^2 y+yz^2+y^2 z+xz^2+x^2 z=x^2 y+xyz+zx^2+xy^2+

                                                                                     y^2 z+xyz+xyz+ yz^2+ xz^2
simplifying -
x^3+y^3+z^3=3xyz --Ans

x^2+y^2+z^2=xy+yz+zx      ----eq(1)

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

x^3+y^3+z^3 -3xyz =(x+y+z)(xy+yz+zx-xy-yz-zx)   (acc. to eq1.)

Therefore , x^3+y^3+z^3 - 3xyz = 0

            So,  x^3+y^3+z^3= 3xyz    Answer 

We will work with the RHS part here,
(x + y+ z)( x² +y² +z² - xy - yz - zx)
Expanding the entire bracket,
= x³ + xy² + xz² - x²y - xyz -zx² + yx + y³ + yz² - xy² - y²z - xyz + zx² + zy² + z³ - xyz - yz² - z²x
=x³ + y³ +z³ - 3xyz
This is equal to LHS
Hence proved

As we know

(X^3 +Y^3+Z^3)-3XYZ = (X+Y+Z) (X^2 +Y^2 +Z^2-XY-YZ-ZX)

given  

X^2 +Y^2 +Z^2=XY+YZ+ZX

I.e X^2 +Y^2 +Z^2-XY-YZ-ZX=0

so

(X^3 +Y^3+Z^3)-3XYZ = (X+Y+Z) (0)

(X^3 +Y^3+Z^3)-3XYZ=0

(X^3 +Y^3+Z^3)=3XYZ

So value of (X^3 +Y^3+Z^3) is (3XYZ) 

Given: x² +y² +z² = xy +yz +zx

We know that

x³ +y³ +z³ −3xyz = (x +y +z)(x² +y² +z² −xy −yz −zx).

Substituting the value of x² +y² +z² in above equation, we get

x³ +y³ +z³ −3xyz = (x +y +z)( xy +yz +zx −xy −yz −zx)

x³ +y³ +z³ −3xyz = (x +y +z) (0) ……  ………………… ( xy +yz +zx −xy −yz –zx ) = 0

x³ +y³ +z³ −3xyz = 0 …………………………………………………….(x +y +z) (0) = 0

x³ +y³ +z³ = 3xyz

Start from Right Hand Side,
(x + y+ z)( x² +y² +z² - xy - yz - zx)
Expand it now,
= x³ + xy² + xz² - x²y - xyz -zx² + yx + y³ + yz² - xy² - y²z - xyz + zx² + zy² + z³ - xyz - yz² - z²x
=x³ + y³ +z³ - 3xyz
equal to LHS
and proved