LA2.1.18

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Show that {\begin{vmatrix}b+c&a-b&a\\c+a&b-c&b\\a+b&c-a&c\end{vmatrix}}=3abc-a^{3}-b^{3}-c^{3}\,

Applying C_{1}+C_{3},C_{2}-C_{3}\, we get

{\begin{vmatrix}a+b+c&-b&a\\a+b+c&-c&b\\a+b+c&-a&c\end{vmatrix}}\,

(a+b+c)(-1){\begin{vmatrix}1&b&a\\1&c&b\\1&a&c\end{vmatrix}}\,

Applying R_{2}-R_{1},C_{2}-C_{3}\, we have

-(a+b+c){\begin{vmatrix}1&b&a\\0&c-b&b-a\\0&a-b&c-a\end{vmatrix}}\,

-(a+b+c)[(c-b)(c-a)-(a-b)(b-a)]=-(a+b+c)[a^{2}+b^{2}+c^{2}-ab-bc-ca]=-(a^{3}+b^{3}+c^{3}-3abc)=3abc-a^{3}-b^{3}-c^{3}\,=RHS


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