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Question Number 56510 by Tawa1 last updated on 17/Mar/19

Commented by MJS last updated on 17/Mar/19

$well, the determinant is$ $2 a^{4} b c + 2 {a b}^{4} c + 2 {a b c}^{4} + 6 a^{3} b^{2} c + 6 a^{3} {b c}^{2} + 6 a^{2} b^{3} c + 6 a^{2} {b c}^{3} + 6 {a b}^{3} c^{2} + 6 {a b}^{2} c^{3} + 12 a^{2} b^{2} c^{2} =$ $= 2 a b c (a^{3} + b^{3} + c^{3} + 3 (a^{2} b + a^{2} c + {a b}^{2} + {a c}^{2} + b^{2} c + {b c}^{2}) + 6 a b c) =$ $= 2 a b c {(a + b + c)}^{3}$
Commented by Tawa1 last updated on 18/Mar/19

$God bless you sir$
	Answered by tanmay.chaudhury50@gmail.com last updated on 18/Mar/19

	$t r y i n g t o s o l v e b y a n o t h e r m e t h o d . . .$ $d e t e r m i n a n t △ h a s d e g r e e = 6$ $1) i f y o u p u t a = 0 i n △$ $∣ {(b + c)}^{2} 0 0 ∣$ $∣ b^{2} c^{2} b^{2} ∣ \leftarrow i t s v a l u e i s 0$ $∣ c^{2} c^{2} b^{2} ∣$ $s o a i s a f a c t o r o f △ . s i m i l a r l y b i s a l s o$ $a f a c t o r o f △ a n d c a l s o a f a c t o r .$ $s o a b c i s f s c t o r o f △$ $n o w p u t c o n d i t i o n a + b + c = 0$ $∣ {(- a)}^{2} a^{2} a^{2} ∣$ $∣ b^{2} {(- b)}^{2} b^{2} ∣$ $∣ c^{2} c^{2} {(- c)}^{2} ∣$ $t h e v a l u e o f t h i s d e t e r m i n a n t = 0$ $h e r e l o o k t h r e e c o l o u m n a r e i d e n t i c a l$ $s o {(a + b + c)}^{3 - 1} \to {(a + b + c)}^{2} i s f a c t o r o f$ $n o w d e g r e e o f (a b c) {(a + b + c)}^{2} i s 5$ $b u t t h e d e g r e e o f △ = 6$ $s o △ = K a b c {(a + b + c)}^{2} \times (a + b + c)$ $[r e d c o l o u r e d (a + b + c) i s s y m m e t r i c f u n c t i o n$ $o f d e g r e e = 1$ $a^{2} + b^{2} + c^{2} s n d a b + b c + c a s y m m e t r i c g u n c t i o n$ $o f d e g r e r = 2 e t c]$ $s o △ = k a b c {(a + b + c)}^{2} (a + b + c)$ $△ =∣ {(b + c)}^{2} a^{2} a^{2} ∣$ $∣ b^{2} {(a + c)}^{2} b^{2} ∣$ $∣ c^{2} c^{2} {(a + b)}^{2} ∣$ $= k a b c {(a + b + c)}^{3}$ $n o w p u t a = b = c = 1 b o t h s i d e . .$ $∣ 4 1 1 ∣$ $∣ 1 4 1 ∣ = k \times 1 \times 1 \times 1 \times 27$ $∣ 1 1 4 ∣$ $54 = 27 k s o k = 2$ $h e n c e a n s w e r i s 2 a b c {(a + b + c)}^{3}$
	Commented by Tawa1 last updated on 18/Mar/19

	$God bless you sir$
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