Deterministic Factorization

From Newton's Sums:

$a^3 + b^3+ c^3 = (a+b+c)(a^2+b^2+c^2) -(a+b+c)(ab + ac + bc) + 3abc$

Therefore,

$a^3 + b^3 + c^3 - 3abc\\ = (a+b+c)(a^2+b^2+c^2) - (a+b+c)(ab+ac+bc) + 3abc - 3abc\\ = (a+b+c)\left((a^2+b^2+c^2) - (ab+ac+bc)\right)\\ =\boxed{(a+b+c)(a^2+b^2+c^2 - ac - ab - cb)}$

First thing to do is realize that $a*a*a + b*b*b + c*c*c -abc -abc -abc$ then is most likely that it will help us to think on this as a determinant $\det \begin{vmatrix} a & b & c\\ c & a & b\\ b & c & a \end{vmatrix}$ , and using determinant properties $\det \begin{vmatrix} a & b & c\\ c & a & b\\ b & c & a \end{vmatrix} = \det \begin{vmatrix} a+b+c & b & c\\ a+b+c & a & b\\ a+b+c & c & a \end{vmatrix}$ , which have a common column, take it off from the det, $(a+b+c)* \det \begin{vmatrix} 1 & b & c\\ 1 & a & b\\ 1 & c & a \end{vmatrix}$ , with some calculations $(a+b+c)* \det \begin{vmatrix} 1 & b & c\\ 1 & a & b\\ 1 & c & a \end{vmatrix} = (a+b+c)( a ^{2} + b^{2} + c^{2} - ac - ab -cb)$