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Question Number 65087 by Tawa1 last updated on 25/Jul/19

	Answered by MJS last updated on 25/Jul/19
	$easier than it seems (1) x+y+z=0 ⇒ z=−x−y insert this in (2) and (3) (2) −3xy(x+y)=18 (3) −7xy(x+y)(x^2 +xy+y^2 )^2 =2058 now simply divide (2) by (−3) (2) xy(x+y)=−6 insert this in (3) (3) 42(x^2 +xy+y^2 )=2058 (x^2 +xy+y^2 )=49 x^2 +xy+y^2 =7 xy+y^2 =7−x^2 x+y=((7−x^2 )/y) now remember xy(x+y)=−6 ⇒ x+y=−(6/(xy)) −(6/(xy))=((7−x^2 )/y) x^3 −7x−6=0 ⇒ x=−2∨x=−1∨x=3 because in all given equations x, y, z are interchangeable we get ((x),(y),(z) ) ∈{ (((−2)),((−1)),(3) ) , (((−2)),(3),((−1)) ) , (((−1)),((−2)),(3) ) , (((−1)),(3),((−2)) ) , ((3),((−2)),((−1)) ) , ((3),((−1)),((−2)) ) }$
	$easier than it seems$ $(1) x + y + z = 0 \Rightarrow z = - x - y$ $insert this in (2) and (3)$ $(2) - 3 x y (x + y) = 18$ $(3) - 7 x y (x + y) {(x^{2} + x y + y^{2})}^{2} = 2058$ $now simply divide (2) by (- 3)$ $(2) x y (x + y) = - 6$ $insert this in (3)$ $(3) 42 (x^{2} + x y + y^{2}) = 2058$ $(x^{2} + x y + y^{2}) = 49$ $x^{2} + x y + y^{2} = 7$ $x y + y^{2} = 7 - x^{2}$ $x + y = \frac{7 - x^{2}}{y}$ $now remember x y (x + y) = - 6 \Rightarrow x + y = - \frac{6}{x y}$ $- \frac{6}{x y} = \frac{7 - x^{2}}{y}$ $x^{3} - 7 x - 6 = 0$ $\Rightarrow x = - 2 \lor x = - 1 \lor x = 3$ $because in all given equations x, y, z are$ $interchangeable we get$ $(\begin{matrix} x \\ y \\ z \end{matrix}) \in {(\begin{matrix} - 2 \\ - 1 \\ 3 \end{matrix}), (\begin{matrix} - 2 \\ 3 \\ - 1 \end{matrix}), (\begin{matrix} - 1 \\ - 2 \\ 3 \end{matrix}), (\begin{matrix} - 1 \\ 3 \\ - 2 \end{matrix}), (\begin{matrix} 3 \\ - 2 \\ - 1 \end{matrix}), (\begin{matrix} 3 \\ - 1 \\ - 2 \end{matrix})}$
	Commented by Tawa1 last updated on 25/Jul/19

	$God bless you sir$
	Commented by Tawa1 last updated on 25/Jul/19

	$Sir i {don}^{'} t understand from line 2 and 3 only .$
	Commented by MJS last updated on 25/Jul/19

	$I inserted some explanations$
	Commented by Tawa1 last updated on 25/Jul/19

	$Ohh, God bless you sir$
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