if (a+bω+cω^2 )+(aω+bω^2 +c)^2 +(aω^2 +b+cω)^2 =0 then prove that a=c or a+c=2b


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Question Number 42009 by 737954 last updated on 16/Aug/18

$if (a + b ω + c ω^{2}) + {(a ω + b ω^{2} + c)}^{2} + {(a ω^{2} + b + c ω)}^{2} = 0$ $then prove that a = c or a + c = 2 b$
	Answered by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18

	$a^{2} + a^{2} w^{2} + a^{2} w^{4} + b^{2} w^{2} + b^{2} w^{4} + b^{2} + c^{2} w^{4} + c^{2} + c^{2} w^{2}$ $+ 2 a b w + 2 {a b w}^{3} + 2 {a b w}^{2} + 2 {b c w}^{3} + 2 {b c w}^{2} + 2 b c w$ $+ 2 {a c w}^{2} + 2 a c w + 2 {a c w}^{3}$ $= a^{2} (1 + w^{2} + w) + b^{2} (w^{2} + w + 1) + c^{2} (w + 1 + w^{2}) +$ $2 a b (w + w^{3} + w^{2}) + 2 b c (w^{3} + w^{2} + w) + 2 a c (w^{2} + w + w^{3})$ $a \times 0 + b \times 0 + c \times 0 + (1 + w + w^{2}) (2 a b + 2 b c + 2 a c)$ $= 0 \times (2 a b + 2 b c + 2 a c$ $= 0$ $p l s c h e c k t h e q u e s t i o n$ $b e c a u s e g i v e n e x p r e s s i o n = 0 o n s i m p l i f i c a t i o n$ $b u t c a n n o t p r o v e a = c o r a + c = 2 b$ $w^{4} = w^{3} . w = 1 . w = w$ $1 + w + w^{2} = 0 f o r m u l a$ $i h a v e s q u a r e d a n d t h e n a d d e d i d e n t i c a l t e r m s$ $b y m i n d c a l c u l a t i o n$ $b e c a u z e$ ${(A + B + C)}^{2} = A^{2} + B^{} + C^{2} + 2 A B + 2 B C + 2 A C$
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