a+b+c=1 a^2 +b^2 +c^2 =2 a^3 +b^3 +c^3 =3 then a^5 +b^5 +c^5 ?


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Question Number 174514 by Best1 last updated on 03/Aug/22

$a + b + c = 1$ $a^{2} + b^{2} + c^{2} = 2$ $a^{3} + b^{3} + c^{3} = 3$ $t h e n a^{5} + b^{5} + c^{5} ?$
Commented by mr W last updated on 03/Aug/22

$s e e Q 74970$
	Answered by RedMath last updated on 03/Aug/22

	$n o i d e a l o l$
	Commented by Best1 last updated on 03/Aug/22

	$s o r r y$
	Answered by MJS_new last updated on 03/Aug/22

	$1 . a + b + c = α$ $2 . a^{2} + b^{2} + c^{2} = β$ $3 . a^{3} + b^{3} + c^{3} = γ$ $========================$ $let a = u - \sqrt{v} \land b = u + \sqrt{v}$ $\Rightarrow$ $1 . c = α - 2 u$ $2 . 6 u^{2} - 4 α u + 2 v + α^{2} - β = 0$ $3 . - 6 u^{3} + 12 α u^{2} + 6 u (v - α^{2}) + α^{3} - γ = 0$ $========================$ $2 . v = - 3 u^{2} + 2 α u - \frac{α^{2} - β}{2}$ $3 . u^{3} - α u^{2} + \frac{3 α^{2} - β}{8} u = \frac{α^{3} - γ}{24}$ $========================$ $a^{4} + b^{4} + c^{4} =$ $= - 32 α (u^{3} - α u^{2} + \frac{3 α^{2} - β}{8} u) + \frac{3 α^{4} - 2 α^{2} β + β^{2}}{2} =$ $= \frac{α^{4} - 6 α^{2} β + 8 α γ + 3 β^{2}}{6}$ $========================$ $a^{5} + b^{5} + c^{5} =$ $= - 20 (α^{2} + β) (u^{3} - α u^{2} + \frac{3 α^{2} - β}{8} u) + α^{5} =$ $= \frac{α^{5} - 5 (α^{3} β - α^{2} γ - 5 l β γ)}{6}$ $========================$
	Commented by Tawa11 last updated on 03/Aug/22

	$Great sirs .$
	Answered by behi834171 last updated on 03/Aug/22

	$a^{2} + a b + a c = a$ $a b + b^{2} + b c = b$ $a c + b c + c^{2} = c \Rightarrow 2 Σ a b = Σ a - Σ a^{2} = 1 - 2 = - 1$ $\Rightarrow Σ a b = - \frac{1}{2}$ $a^{3} + a (b^{2} + c^{2}) = 2 a$ $b (a^{2} + c^{2}) + b^{3} = 2 b$ $c (a^{2} + b^{2}) + c^{3} = 2 c \Rightarrow$ $\Rightarrow Σ a b (a + b) = 2 Σ a - Σ a^{3} = 2 - 3 = - 1$ $\Rightarrow Σ a b (1 - c) = - 1 \Rightarrow Σ a b - 3 a b c = - 1$ $\Rightarrow - 3 a b c = - 1 + \frac{1}{2} = - \frac{1}{2} \Rightarrow a b c = \frac{1}{6}$ $s o : a, b, c, a r e t h e r o o t s o f e q u a t i o n :$ $z^{3} - z^{2} - \frac{1}{2} z - \frac{1}{6} = 0$ $z^{5} = z^{3} . z^{2} = z^{2} (z^{2} + \frac{1}{2} z + \frac{1}{6}) = z^{4} + \frac{1}{2} z^{3} + \frac{1}{6} z^{2} =$ $= z (z^{2} + \frac{1}{2} z + \frac{1}{6}) + \frac{1}{2} (z^{2} + \frac{1}{2} z + \frac{1}{6}) + \frac{1}{6} z^{2} =$ $= z^{2} + \frac{1}{2} z + \frac{1}{6} + \frac{1}{2} z^{2} + \frac{1}{6} z + \frac{1}{2} z^{2} + \frac{1}{4} z + \frac{1}{12} + \frac{1}{6} z^{2} =$ $= \frac{13}{6} z^{2} + \frac{11}{12} z + \frac{1}{4}$ $\Rightarrow Σ a^{5} = \frac{13}{6} Σ a^{2} + \frac{11}{12} Σ a + \frac{3}{4} =$ $= \frac{13}{6} \times 2 + \frac{11}{12} \times 1 + \frac{3}{4} = 6 . ◼$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$ $b y u s i n g s y m b o l s o f d e a r m r : M J S_n e w$ $Σ a b = \frac{Σ a - Σ a^{2}}{2} = \frac{α - β}{2}$ $Σ a b - 3 a b c = 2 Σ a - Σ a^{3} \Rightarrow$ $a b c = - \frac{1}{3} (2 α - γ - \frac{α - β}{2}) = - \frac{3 α + β - 2 γ}{6}$ $\Rightarrow z^{3} - α z^{2} + \frac{α - β}{2} z + \frac{3 α + β - 2 γ}{6} = 0$ $\Rightarrow z^{5} = z^{2} z^{3} = z^{2} (α z^{2} - \frac{α - β}{2} z - \frac{3 α + β - 2 γ}{6}) =$ $= α z (α z^{2} - \frac{α - β}{2} z - \frac{3 α + β - 2 γ}{6}) - \frac{α - β}{2} (α z^{2} - \frac{α - β}{2} z - \frac{3 α + β - 2 γ}{6}) - \frac{3 α + β - 2 γ}{6} z^{2} =$ $= α^{2} (α z^{2} - \frac{α - β}{2} z - \frac{3 α + β - 2 γ}{6}) - \frac{α (α - β)}{2} z^{2} - \frac{α (3 α + β - 2 γ)}{6} z -$ $+ \frac{α (α - β)}{2} z^{2} + \frac{{(α - β)}^{2}}{4} z + \frac{(α - β) (3 α + β - 2 γ)}{12} - \frac{3 α + β - 2 γ}{6} z^{2} =$ $= [α^{3} - \frac{α (α - β)}{2} - \frac{3 α + β - 2 γ}{6}] z^{2} - [\frac{α^{2} (α - β)}{2} + \frac{α (3 α + β - 2 γ)}{6} - \frac{{(α - β)}^{2}}{4}] . z -$ $- \frac{(2 α^{2} - α + β) (3 α + β - 2 γ)}{12}$ $= A . z^{2} + B . z + C$ $A = \frac{6 α^{3} - 3 α^{2} + 3 α β - 3 α - β + 2 γ}{6}$ $B = - \frac{6 α^{3} - 6 α^{2} β + 3 α^{2} + 8 α β - 4 α γ - 3 β^{2}}{12}$ $C = - \frac{6 α^{3} + 2 α^{2} β - 4 α^{2} γ - 3 α^{2} + 2 α β + 2 α γ + β^{2} - 2 β γ}{12}$ $\Rightarrow Σ a^{5} = A . Σ a^{2} + B . Σ a + 3 C$
	Commented by Best1 last updated on 03/Aug/22

	$w a w w w t h a n k s g u y s$
	Commented by Tawa11 last updated on 03/Aug/22

	$Great sirs$
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