Question

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

Answer 1

Given $\left|\begin{array}{ccc}a& a^3& a^4-1\\ b& b^3& b^4-1\\ c& c^3& c^4-1\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}a& a^3& a^4\\ b& b^3& b^4\\ c& c^3& c^4\end{array}\right|+\left|\begin{array}{ccc}a& a^3& -1\\ b& b^3& -1\\ c& c^3& -1\end{array}\right|=0$
$\Rightarrow abc\left|\begin{array}{ccc}1& a^2& a^3\\ 1& b^2& b^3\\ 1& c^2& c^3\end{array}\right|-\left|\begin{array}{ccc}a& a^3& 1\\ b& b^3& 1\\ c& c^3& 1\end{array}\right|=0$
$\Rightarrow abc(a-c)(b-c)(c-a)(ab+b{c}_{c}a)-(a-b)(b-c)(c-a)(a+b+c)=0$
$\Rightarrow abc(ab+bc+ca)-(a+b+c)=0$ $[\because a\ne b\ne c]$
$\Rightarrow abc(ab+bc+ca)=a+b+c$

Hence proved