65. Resolve into factors: \( \left(\sum_{x, y, z} x\right)^{3}-\sum_{x, y, z} x^{3} \) (a) \( (x+y)(y+z)(z+x) \) (b) \( -(x+y)(y+z)(z+x) \) (c) \( 3(x+y)(y+z)(z+x) \) (d) \( -3(x+y)(y+z)(z+x) \)

To resolve the expression \( \left(\sum_{x, y, z} x\right)^{3}-\sum_{x, y, z} x^{3} \), we start by rewriting it in a more manageable form. Let \( S = x + y + z \). Then we can express the first term as: \[ S^3 = (x + y + z)^3 \] Using the binomial expansion, we have: \[ S^3 = x^3 + y^3 + z^3 + 3(x+y)(y+z)(z+x) \] Now, we need to compute \( \sum_{x, y, z} x^3 \). This is simply: \[ x^3 + y^3 + z^3 \] Now we can substitute these into our original expression: \[ S^3 - \sum_{x, y, z} x^3 = (x^3 + y^3 + z^3 + 3(x+y)(y+z)(z+x)) - (x^3 + y^3 + z^3) \] This simplifies to: \[ 3(x+y)(y+z)(z+x) \] Thus, we have: \[ \left(\sum_{x, y, z} x\right)^{3}-\sum_{x, y, z} x^{3} = 3(x+y)(y+z)(z+x) \] Therefore, the correct answer is: \[ \boxed{3(x+y)(y+z)(z+x)} \] This corresponds to option (c).