does any one know how to factor (x + y + z) (yz + zx + xy) - xyz?

Method 1 :

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

Put x = -y :
Expression = [(-y) + y + z][yz + z(-y) + (-y)y] - (-y)yz = 0
Hence, (x + y) is a factor of the expreesion.
Also, (y + z) and (z + x) are also factors of the expression.

The expression in order 3, thus
(x + y + z) (yz + zx + xy) - xyz = k(x + y)(y + z)(z +x)

Put x = y = z = 1 :
(1 + 1 + 1)(1 + 1 + 1) - 1 = k(1 + 1)(1 + 1)(1 + 1)
k = 8
k = 1

Hence, (x + y + z)(yz + zx + xy) - xyz = (x + y)(y + z)(z +x)


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Method 2 :

(x + y + z)(yz + zx + xy) - xyz
= x(yz + zx + xy) + y(yz + zx + xy) + z(yz + zx + xy) - xyz
= xyz + zx² + x²y + y²z + xyz + xy² + yz² + z²x + xyz - xyz
= (xyz + x²y) + (z²x + zx²) + (y²z + xy²) + (yz² + xyz)
= xy(z + x) + zx(z + x) + y²(z + x) + yz(z + x)
= (xy + zx)(z + x) + (y² + yz)(z + x)
= x(y + z)(z + x) + y(y + z)(z + x)
= (x + y)(y + z)(z + x)

Set x = -y. Expression = z(xy) - xyz = 0.
Similarly y = -z and z = -x give zero. So
x + y, y + z and z + z are factors. Hence
(x + y + z) (yz + zx + xy) - xyz = k(x + y)(y + z)(z + x).
Setting x = y = z = 1 gives k = 1.
Hence (x + y + z) (yz + zx + xy) - xyz = (x + y)(y + z)(z + x).

This was answered earlier. Expand first then factor. The result is
(x + y)(y + z)(z + x)