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Question Number 35319 by Tinkutara last updated on 17/May/18

	Answered by MJS last updated on 17/May/18

	${it}^{'} s just multiplicating and factorising$ $\det \| \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix} \| = a e i + b f g + c d h - a f h - b d i - c e g$
	Commented by Tinkutara last updated on 18/May/18
	No it is to be done by properties of determinants. Please can you try that way?
	Answered by ajfour last updated on 18/May/18
	$determinant ((((b+c)^2 ),a^2 ,a^2 ),(b^2 ,((c+a)^2 ),b^2 ),(c^2 ,c^2 ,((a+b)^2 ))) C_1 →C_1 −C_3 ; C_2 →C_2 −C_3 = determinant ((((b+c+a)(b+c−a)),0,a^2 ),(0,((c+a+b)(c+a−b)),b^2 ),(((c+a+b)(c−a−b)),((c+a+b)(c−a−b)),((a+b)^2 ))) =(a+b+c)^2 determinant (((b+c−a),0,a^2 ),(0,(c+a−b),b^2 ),((c−a−b),(c−a−b),((a+b)^2 ))) R_3 →R_3 −(R_1 +R_2 ) =(a+b+c)^2 determinant (((b+c−a),0,a^2 ),(0,(c+a−b),b^2 ),((−2b),(−2a),(2ab))) =(a+b+c)^2 {(b+c−a)[2abc+2a^2 b] +a^2 [2bc+2ab−2b^2 ]} =(a+b+c)^2 {2b[abc+ac^2 −a^2 c a^2 b+a^2 c−a^3 +a^2 c+a^3 −a^2 b]} =(a+b+c)^2 {2b[abc+ac^2 +a^2 c} =2abc(a+b+c)^2 (b+c+a) =2abc(a+b+c)^3 .$
	$\| \begin{matrix} {(b + c)}^{2} & a^{2} & a^{2} \\ b^{2} & {(c + a)}^{2} & b^{2} \\ c^{2} & c^{2} & {(a + b)}^{2} \end{matrix} \|$ $C_{1} \to C_{1} - C_{3}; C_{2} \to C_{2} - C_{3}$ $= \| \begin{matrix} (b + c + a) (b + c - a) & 0 & a^{2} \\ 0 & (c + a + b) (c + a - b) & b^{2} \\ (c + a + b) (c - a - b) & (c + a + b) (c - a - b) & {(a + b)}^{2} \end{matrix} \|$ $= {(a + b + c)}^{2} \| \begin{matrix} b + c - a & 0 & a^{2} \\ 0 & c + a - b & b^{2} \\ c - a - b & c - a - b & {(a + b)}^{2} \end{matrix} \|$ $R_{3} \to R_{3} - (R_{1} + R_{2})$ $= {(a + b + c)}^{2} \| \begin{matrix} b + c - a & 0 & a^{2} \\ 0 & c + a - b & b^{2} \\ - 2 b & - 2 a & 2 a b \end{matrix} \|$ $= {(a + b + c)}^{2} {(b + c - a) [2 a b c + 2 a^{2} b]$ $+ a^{2} [2 b c + 2 a b - 2 b^{2}]}$ $= {(a + b + c)}^{2} {2 b [a b c + {a c}^{2} - a^{2} c$ $a^{2} b + a^{2} c - a^{3} + a^{2} c + a^{3} - a^{2} b]}$ $= {(a + b + c)}^{2} {2 b [a b c + {a c}^{2} + a^{2} c}$ $= 2 a b c {(a + b + c)}^{2} (b + c + a)$ $= 2 a b c {(a + b + c)}^{3} .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 18/May/18

	$p l s l e t m e k n o w h o w t o p u t d e t e r m i n a n t s i g n$ $i n t h i s a p p$
	Commented by ajfour last updated on 18/May/18

	$i {c o u l d n}^{'} t f i n d a w a y l i k e t h a t . .$
	Commented by Tinkutara last updated on 18/May/18

	$S h o u l d n o t w e s i m p l i f y t h e d e t e r m i n a n t$ $t i l l a t l e a s t 2 z e r o e s i n a n y r o w o r$ $c o l u m n ?$
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