a+b+c=1 a^2 +b^2 +c^2 =1 a^3 +b^3 +c^3 =1 a , b , c =?


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Question Number 147116 by Rasheed.Sindhi last updated on 18/Jul/21

$a + b + c = 1$ $a^{2} + b^{2} + c^{2} = 1$ $a^{3} + b^{3} + c^{3} = 1$ $a, b, c = ?$
Commented by mr W last updated on 18/Jul/21

$(a, b, c) = (0, 0, 1)$ $w e c a n g e t a b + b c + c a = 0, a b c = 0,$ $t h e r e f o r e a, b, c a r e r o o t s o f$ $x^{3} - x^{2} + 0 x + 0 = 0 \Rightarrow x^{2} (x - 1) = 0$ $\Rightarrow x \in (0, 0, 1)$
Commented by Rasheed.Sindhi last updated on 18/Jul/21

$S i r I^{'} v e t r i e d t o s o l v e i n a d i f f e r e n t$ $w a y i n w h i c h 2 a l s o a p p e a r s a s a$ $r o o t . I k n o w t h i s {c a n}^{'} t s a t i s f y t h e$ $s y s t e m . B u t t h e n I {c a n}^{'} t d e s i d e$ $w h e t h e r {i t}^{'} s e x t r a n e o u s r o o t b e c a u s e$ $I {h a v e n}^{'} t s q u a r e d a n y e q u a t i o n .$ $P l e a s e s e e m y s o l u t i o n a n d g u i d e$ $m e a s y o u a l w a y s d o .$
Commented by Rasheed.Sindhi last updated on 18/Jul/21

$S o r r y s i r I f o u n d m y e r r o r w h e n$ $I s o l v e d i t t h i s t i m e!$ $T h a n k s S i r .$
Commented by mr W last updated on 18/Jul/21

$i t h i n k y o u a l s o g e t t h e r i g h t a n s w e r$ $i n t h a t w a y . y o u g e t c = 1 t h e n a + b = 0$ $a n d a b = 0, i . e . a = b = 0 . o r c = 0 t h e n$ $a + b = 1 a n d a b = 0, i . e . a = 1, b = 0 o r$ $a = 0, b = 1 .$
Commented by Rasheed.Sindhi last updated on 18/Jul/21

$T h a n X f o r t h i s n i c e t r i c k S i r,$ $I {d i d n}^{'} t t h i n k o f t h a t .$
	Answered by Olaf_Thorendsen last updated on 18/Jul/21
	$a+b+c = 1 a^2 +b^2 +c^2 = 1 a^3 +b^3 +c^3 = 1 a^2 +b^2 +c^2 = (a+b+c)^2 −2(ab+bc+ca) 1 = 1−2(ab+bc+ca) ab+bc+ca = 0 a^3 +b^3 +c^3 = (a+b+c)(a^2 +b^2 +c^2 ) −(ab^2 +ac^2 +a^2 b+bc^2 +a^2 c+b^2 c) 1 = 1×1−(ab^2 +ac^2 +a^2 b+bc^2 +a^2 c+b^2 c) 0 = a(ab+ca)+b(ab+bc)+c(bc+ca) 0 = a(−bc)+b(−ca)+c(−ab) abc = 0 we verify that a = 0 ⇒ b = 0 and c = 1 or b = 1 and c = 0 S = {(1,0,0) ; (0,1,0) ; (0,0,1)}$
	$a + b + c = 1$ $a^{2} + b^{2} + c^{2} = 1$ $a^{3} + b^{3} + c^{3} = 1$ $a^{2} + b^{2} + c^{2} = {(a + b + c)}^{2} - 2 (a b + b c + c a)$ $1 = 1 - 2 (a b + b c + c a)$ $a b + b c + c a = 0$ $a^{3} + b^{3} + c^{3} = (a + b + c) (a^{2} + b^{2} + c^{2})$ $- ({a b}^{2} + {a c}^{2} + a^{2} b + {b c}^{2} + a^{2} c + b^{2} c)$ $1 = 1 \times 1 - ({a b}^{2} + {a c}^{2} + a^{2} b + {b c}^{2} + a^{2} c + b^{2} c)$ $0 = a (a b + c a) + b (a b + b c) + c (b c + c a)$ $0 = a (- b c) + b (- c a) + c (- a b)$ $a b c = 0$ $w e v e r i f y t h a t a = 0 \Rightarrow b = 0 and c = 1$ $or b = 1 and c = 0$ $S = {(1, 0, 0); (0, 1, 0); (0, 0, 1)}$
	Commented by Rasheed.Sindhi last updated on 18/Jul/21

	$T h a n k s S i r Olaf! I a l w a y s a p p r e c i a t e$ $y o u r p r e c i o u s c o n t r i b u t i o n t o w a r d s t h i s$ $f o r u m!$
	Answered by Rasheed.Sindhi last updated on 18/Jul/21
	$a+b+c=a^2 +b^2 +c^2 =a^3 +b^3 +c^3 =1_(−) ;a,b,c=? (i)⇒a+b=1−c (ii)⇒(a+b)^2 −2ab+c^2 −1=0 (1−c)^2 −2ab+c^2 −1=0 ab=((1−2c+c^2 +c^2 −1)/2)=c^2 −c...A (iii)⇒(a+b)^3 −3ab(a+b)+c^3 −1=0 (1−c)^3 −3ab(1−c)−(1−c^3 )=0 (1−c){(1−c)^2 −3ab−(1+c+c^2 )}=0 c=1 ∨ 1−2c+c^2 −3ab−1−c−c^2 =0 ab=−c...........................B A & B : c^2 −c=−c c^2 =0⇒c=0 Observing symmetric nature of the system we can fix same values for a & b. • •^(•) (a,b,c)=(1,0,0) , (0,1,0) , (0,0,1)$
	$\underline{a + b + c = a^{2} + b^{2} + c^{2} = a^{3} + b^{3} + c^{3} = 1}; a, b, c = ?$ $(i) \Rightarrow a + b = 1 - c$ $(i i) \Rightarrow {(a + b)}^{2} - 2 a b + c^{2} - 1 = 0$ ${(1 - c)}^{2} - 2 a b + c^{2} - 1 = 0$ $a b = \frac{1 - 2 c + c^{2} + c^{2} - 1}{2} = c^{2} - c . . . A$ $(i i i) \Rightarrow {(a + b)}^{3} - 3 a b (a + b) + c^{3} - 1 = 0$ ${(1 - c)}^{3} - 3 a b (1 - c) - (1 - c^{3}) = 0$ $(1 - c) {{(1 - c)}^{2} - 3 a b - (1 + c + c^{2})} = 0$ $c = 1 \lor 1 - 2 c + c^{2} - 3 a b - 1 - c - c^{2} = 0$ $a b = - c . . . . . . . . . . . . . . . . . . . . . . . . . . . B$ $A & B : c^{2} - c = - c$ $c^{2} = 0 \Rightarrow c = 0$ $O b s e r v i n g s y m m e t r i c n a t u r e o f$ $t h e s y s t e m w e c a n f i x s a m e v a l u e s$ $f o r a & b .$ $\overset{∙}{∙ ∙} (a, b, c) = (1, 0, 0), (0, 1, 0), (0, 0, 1)$
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