If $\left|\begin{array}{ccc}a& a^3& a^4\\ b& b^3& b^4\\ c& c^3& c^4\end{array}\right|$$=k(a-b)(b-c)(c-a) then k=$

$$\begin{gathered} \left|\begin{array}{ccc}a& a^3& a^4\\ b& b^3& b^4\\ c& c^3& c^4\end{array}\right| =abc\left|\begin{array}{ccc}1& a^2& a^3\\ 1& b^2& b^3\\ 1& c^2& c^3\end{array}\right| \\ {R}_2\to R_2-R_1,R_3\to R_3-R_1 \\ =abc\left|\begin{array}{ccc}1& a^2& a^3\\ 0& b^2-a^2& b^3-a^3\\ 1& c^2-a^2& c^3-a^3\end{array}\right| \\ =abc(b-a)(c-a)\left|\begin{array}{ccc}1& a^2& a^3\\ 0& b+a& b^2+ab+a^2\\ 0& c+a& c^2+ac+a^2\end{array}\right| \\ =abc(b-a)(c-a)\left|\begin{array}{cc}b+a& b^2+ab+a^2\\ c+a& c^2+ac+a^2\end{array}\right| \\ =abc(b-a)(c-a)\left|\begin{array}{cc}b+a& b^2+ab+a^2\\ c-b& c^2-b^2+a\left(c-b\right)\end{array}\right| \\ =abc(b-a)(c-a)(c-b)\left|\begin{array}{cc}b+a& b^2+ab+a^2\\ 1& a+b+c\end{array}\right| \\ =abc(ab + bc + ca)(b-a)(c-a)(c-b) \end{gathered}$$