Resolve the system d′ unknow (x, y,z) ∈ ⊂^3 x+y+z=1 x^2 +y^2 +z^2 =1 x^3 +y^3 +z^3 =−5


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Question Number 158855 by LEKOUMA last updated on 09/Nov/21

$R e s o l v e t h e s y s t e m d^{'} u n k n o w (x, y, z) \in \subset^{3}$ $x + y + z = 1$ $x^{2} + y^{2} + z^{2} = 1$ $x^{3} + y^{3} + z^{3} = - 5$
	Answered by mr W last updated on 09/Nov/21

	$x + y + z = 1$ ${(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 (x y + y z + z x)$ $1^{2} = 1 + 2 (x y + y z + z x)$ $\Rightarrow x y + y z + z x = 0$ ${(x + y + z)}^{3} = x^{3} + y^{3} + z^{3} + 3 (x + y + z) (x y + y z + z x) - 3 x y z$ $1^{3} = - 5 + 3 \times 1 \times 0 - 3 x y z$ $\Rightarrow x y z = - 2$ $x, y, z a r e r o o t s o f$ $t^{3} - t^{2} + 2 = 0$ $(t + 1) (t - 2 t + 2) = 0$ $\Rightarrow (x, y, z) = (- 1, 1 + i, 1 - i)$
	Commented by LEKOUMA last updated on 09/Nov/21

	$M a n y t h a n k s$
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