If \[\left| \begin{array}{l} a\,\,\,\,{a^2}\,\,\,\,\,1 + {a^3}\\ b\,\,\,\,{b^2}\,\,\,\,\,1 + {b^3}\\ c\,\,\,\,\,{c^2}\,\,\,\,\,1 + {c^3} \end{array} \right| = 0\]     and vectors $$\left( {1,\,a,{a^2}} \right),\,\left( {1,\,b,{b^2}} \right)$$     and $$\left( {1,\,c,{c^2}} \right)$$  are non-coplanar, then the product $$abc$$  equals :

Solution :
\[\begin{array}{l} \left| \begin{array}{l} a\,\,\,\,\,{a^2}\,\,\,\,\,1 + {a^3}\\ b\,\,\,\,\,{b^2}\,\,\,\,\,1 + {b^3}\\ c\,\,\,\,\,{c^2}\,\,\,\,\,1 + {c^3} \end{array} \right| = 0\\ \Rightarrow \left| \begin{array}{l} a\,\,\,\,{a^2}\,\,\,\,\,1\\ b\,\,\,\,{b^2}\,\,\,\,\,1\\ c\,\,\,\,\,{c^2}\,\,\,\,\,1 \end{array} \right| + \left| \begin{array}{l} a\,\,\,\,{a^2}\,\,\,\,\,{a^3}\\ b\,\,\,\,{b^2}\,\,\,\,\,{b^3}\\ c\,\,\,\,\,{c^2}\,\,\,\,\,{c^3} \end{array} \right| = 0\\ \Rightarrow \left( {1 + abc} \right)\left| \begin{array}{l} 1\,\,\,\,\,a\,\,\,\,{a^2}\\ 1\,\,\,\,\,b\,\,\,\,{b^2}\\ 1\,\,\,\,\,c\,\,\,\,\,{c^2} \end{array} \right| = 0\\ {\rm{As }}\left| \begin{array}{l} 1\,\,\,\,\,a\,\,\,\,{a^2}\\ 1\,\,\,\,\,b\,\,\,\,{b^2}\\ 1\,\,\,\,\,c\,\,\,\,\,{c^2} \end{array} \right| \ne 0\,\,\,\,\left( {{\rm{given \,\,condition}}} \right)\\ \therefore \,\,abc = - 1 \end{array}\]