LA2.1.12

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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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