If $a,b,c$  are all distinct and $\left|\begin{array}{ccc}a& a^3& a^4-1\\ b& b^3& b^4-1\\ c& c^3& c^4-1\end{array}\right|=0$ , then $abc\left(ab+bc+ca\right)=$

$\left|\begin{array}{ccc}a& a^3& a^4\\ b& b^3& b^4\\ c& c^3& c^4\end{array}\right|-\left|\begin{array}{ccc}a& a^3& 1\\ b& b^3& 1\\ c& c^3& 1\end{array}\right|=0$

$abc\left|\begin{array}{ccc}1& a^2& a^3\\ 1& b^2& b^3\\ 1& c^2& c^3\end{array}\right|+\left|\begin{array}{ccc}a& 1& a^3\\ b& 1& b^3\\ c& 1& c^3\end{array}\right|=0$

$abc\left|\begin{array}{ccc}1& a^2& a^3\\ 1& b^2& b^3\\ 1& c^2& c^3\end{array}\right|+\left|\begin{array}{ccc}1& a& a^3\\ 1& b& b^3\\ 1& c& c^3\end{array}\right|=0$

$abc\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(ab+bc+ca\right)-\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)=0$

$abc\left(ab+bc+ca\right)=a+b+c$