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Question Number 190448 by Abdullahrussell last updated on 03/Apr/23

	Answered by a.lgnaoui last updated on 04/Apr/23
	$a+b+c =6 (1) ab+bc+ca=3 (2) abc =−1 (3) (1)×(2)⇒ a^2 b+abc+a^2 c+ab^2 + +cb^2 +abc+abc+bc^2 +ac^2 =3(abc)+(a^2 b+b^2 c+c^2 a)+ (ab^2 +bc^2 +ca^2 ) 21=(a^2 b+b^2 c+c^2 a)+(ab^2 +bc^2 +ca^2 )(4) (a+b+c)^2 =(a^2 +b^2 +c^2 )+6 a^2 +b^2 +c^2 =30 (5) a^3 +b^3 +c^3 =(a+b+c)^3 − 3[(a+b)+c][c(a+b)] =6^3 −3(6)(ac+ab) a^3 +b^3 +c^3 =216−18bc (6) (a+b+c)(a^2 +b^2 +c^2 ) =a^3 +b^3 +c^3 +ab^2 +ac^2 + ba^2 +bc^2 +ca^2 +cb^2 =216−18bc+21=237−18bc =6×30=180 18bc=237−180 ⇒ bc=((19)/6) abc=−1⇒ a=((−6)/(19)) { ((a=((−6)/(19)) bc=((19)/6))),((a+(b+c)=6)) :} b+c=6+(6/(19))=((120)/(19)) { ((b+c=((120)/(19)))),((bc =((19)/6))) :} z^2 −((120)/(19))z+((19)/6)=0 △=(((60)/(19)))^2 −((19)/6) =((3600×6−19^3 )/(19^2 ×6))= z=(0,55 ; 5,766) (a,b ,c)={(((−6)/(19)) , 0,55, 5,766);(((−6)/(19)),5,766, 0,55)} calcul de: a^2 b+b^2 c+c^2 a ((36)/(19^2 ))×0,55+0,55^2 ×5,766−5,766^2 ×(6/(19)) a^2 b+b^2 c+c^2 a=−8,65$
	$a + b + c = 6 (1)$ $a b + b c + c a = 3 (2)$ $a b c = - 1 (3)$ $(1) \times (2) \Rightarrow a^{2} b + a b c + a^{2} c + {a b}^{2} +$ $+ {c b}^{2} + a b c + a b c + {b c}^{2} + {a c}^{2}$ $= 3 (a b c) + (a^{2} b + b^{2} c + c^{2} a) +$ $({a b}^{2} + {b c}^{2} + {c a}^{2})$ $21 = (a^{2} b + b^{2} c + c^{2} a) + ({a b}^{2} + {b c}^{2} + {c a}^{2}) (4)$ ${(a + b + c)}^{2} = (a^{2} + b^{2} + c^{2}) + 6$ $a^{2} + b^{2} + c^{2} = 30 (5)$ $a^{3} + b^{3} + c^{3} = {(a + b + c)}^{3} -$ $3 [(a + b) + c] [c (a + b)]$ $= 6^{3} - 3 (6) (a c + a b)$ $a^{3} + b^{3} + c^{3} = 216 - 18 b c (6)$ $(a + b + c) (a^{2} + b^{2} + c^{2})$ $= a^{3} + b^{3} + c^{3} + {a b}^{2} + {a c}^{2} +$ ${b a}^{2} + {b c}^{2} + {c a}^{2} + {c b}^{2}$ $= 216 - 18 b c + 21 = 237 - 18 b c$ $= 6 \times 30 = 180$ $18 b c = 237 - 180 \Rightarrow b c = \frac{19}{6}$ $a b c = - 1 \Rightarrow a = \frac{- 6}{19}$ ${\begin{cases} a = \frac{- 6}{19} b c = \frac{19}{6} \\ a + (b + c) = 6 \end{cases}$ $b + c = 6 + \frac{6}{19} = \frac{120}{19}$ ${\begin{cases} b + c = \frac{120}{19} \\ b c = \frac{19}{6} \end{cases}$ $z^{2} - \frac{120}{19} z + \frac{19}{6} = 0 △ = {(\frac{60}{19})}^{2} - \frac{19}{6}$ $= \frac{3600 \times 6 - 19^{3}}{19^{2} \times 6} =$ $z = (0, 55; 5, 766)$ $(a, b, c) = {(\frac{- 6}{19}, 0, 55, 5, 766); (\frac{- 6}{19}, 5, 766, 0, 55)}$ $c a l c u l d e : a^{2} b + b^{2} c + c^{2} a$ $\frac{36}{19^{2}} \times 0, 55 + 0, 55^{2} \times 5, 766 - 5, 766^{2} \times \frac{6}{19}$ $a^{2} b + b^{2} c + c^{2} a = - 8, 65$
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