# LA2.1.12

If a,b,c are distinct, $abc\not\equiv 0 \,$ and $\begin{vmatrix} a & a^3 & a^4-1 \\ b & b^3 & b^4-1 \\ c & c^3 & c^4-1 \end{vmatrix}=0\,$ then prove that $abc(ab+bc+ca)=a+b+c\,$

$\begin{vmatrix} a & a^3 & a^4-1 \\ b & b^3 & b^4-1 \\ c & c^3 & c^4-1 \end{vmatrix}=0\,$

$\begin{vmatrix} a & a^3 & a^4 \\ b & b^3 & b^4 \\ c & c^3 & c^4 \end{vmatrix}-\begin{vmatrix} a & a^3 & 1 \\ b & b^3 & 1 \\ c & c^3 & 1 \end{vmatrix}=0\,$

$abc\begin{vmatrix} 1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3 \end{vmatrix}-\begin{vmatrix} a & a^3 & 1 \\ b-a & b^3-c^3 & 0 \\ c-a & c^3-a^3 & 0 \end{vmatrix}=0\,$

$abc\begin{vmatrix} 1 & a^2 & a^3 \\ 0 & b^2-a^2 & b^3-a^3 \\ c & c^2-a^2 & c^3-a^3 \end{vmatrix}-(b-a)(c-a)\begin{vmatrix} a & a^3 & 1 \\ 1 & b^2+ab+a^2 & 0 \\ 1 & c^2+ca+a^2 & 0 \end{vmatrix}=0\,$

$abc(b-a)(c-a)\begin{vmatrix} 1 & a^2 & a^3 \\ 0 & b+a & b^2+ab+a^2 \\ 0 & c+a & c^2+ca+a^2 \end{vmatrix}-(b-a)(c-a)[(c^2+ca+a^2)-(b^2+ab+a^2)]=0\,$

$abc[(b+a)(c^2+ca+a^2)-(c+a)(b^2+ab+a^2)]-[c^2-b^2+ca-ab]=0\,$

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

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