Question

Factorize
a3 + b3 - c3 + 3abc

Answer 1

$\forall \:a,b,c$

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

Substituting c by (–c)

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

And Over the Complex Numbers:

$\boxed{(a+b-c)(a+b\omega-c{\omega}^{2})(a+b{\omega}^{2}-c{\omega})}$

where,

${\omega}^2 +{\omega} +1=0$