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Question Number 56090 by Tawa1 last updated on 10/Mar/19

Commented by math1967 last updated on 10/Mar/19

$p l s c h e c k I t h i n k i t s h o u l d b e$ $\| \begin{matrix} b c & c a & a b \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{matrix} \|$
Commented by Tawa1 last updated on 10/Mar/19

$Please help me do it like this$
Commented by Tawa1 last updated on 10/Mar/19

$But that is the question .$
Commented by math1967 last updated on 10/Mar/19
$determinant (((bc−ca),(ca−ab),(ab)),((a−b),(b−c),c),((a^2 −b^2 ),(b^2 −c^(2 ) ),(c^(2 ) )))c_(1 ) −c_2 ,c_(2 ) −c_(3 ) (a−b)(b−c) determinant (((−c),(−a),(ab)),(1,1,c),((a+b),(b+c),c^2 )) =(a−b)(b−c) determinant (((−c),(c−a),(ab)),(1,0,c),((a+b ),(c−a),c^2 ))c_2 −c_1 =(a−b)(b−c)(c−a) determinant (((−c),1,(ab)),(1,0,c),((a+b),1,c^2 )) =(a−b)(b−c)(c−a){c^2 −c^2 +ca+bc+ab} =(b−a)(c−b)(c−a)(ab+bc+ca)$
$\| \begin{matrix} b c - c a & c a - a b & a b \\ a - b & b - c & c \\ a^{2} - b^{2} & b^{2} - c^{2} & c^{2} \end{matrix} \| c_{1} - c_{2}, c_{2} - c_{3}$ $(a - b) (b - c) \| \begin{matrix} - c & - a & a b \\ 1 & 1 & c \\ a + b & b + c & c^{2} \end{matrix} \|$ $= (a - b) (b - c) \| \begin{matrix} - c & c - a & a b \\ 1 & 0 & c \\ a + b & c - a & c^{2} \end{matrix} \| c_{2} - c_{1}$ $= (a - b) (b - c) (c - a) \| \begin{matrix} - c & 1 & a b \\ 1 & 0 & c \\ a + b & 1 & c^{2} \end{matrix} \|$ $= (a - b) (b - c) (c - a) {c^{2} - c^{2} + c a + b c + a b}$ $= (b - a) (c - b) (c - a) (a b + b c + c a)$
Commented by Tawa1 last updated on 11/Mar/19

$God bless you sir$
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