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Question Number 158598 by HongKing last updated on 06/Nov/21

Commented by MJS_new last updated on 06/Nov/21

$let y = p x \land z = q x then {it}^{'} s easy to solve . I get$ $x = \sqrt[3]{6} \land y = - \frac{x}{3} \land z = - \frac{x}{2}$
Commented by HongKing last updated on 06/Nov/21

$Yes my dear Ser$ $I am interested in you creative solution$ $if possible please$
	Answered by Rasheed.Sindhi last updated on 06/Nov/21
	$What′s wrong in the following process: { ((−x^2 (y−z)=−1)),((y^2 (z−x)=−1)),((z^2 (x−y)=2)) :} Adding all the three: −x^2 (y−z)+y^2 (z−x)+z^2 (x−y)=0 (y−z)(−x^2 −xy+yz−zx)=0 y−z=0 ∣ −x^2 −xy+yz−zx=0 y=z ???????????????? Obviously false!$
	${W h a t}^{'} s w r o n g i n t h e f o l l o w i n g p r o c e s s :$ ${\begin{cases} - x^{2} (y - z) = - 1 \\ y^{2} (z - x) = - 1 \\ z^{2} (x - y) = 2 \end{cases}$ $A d d i n g a l l t h e t h r e e :$ $- x^{2} (y - z) + y^{2} (z - x) + z^{2} (x - y) = 0$ $(y - z) (- x^{2} - x y + y z - z x) = 0$ $y - z = 0 ∣ - x^{2} - x y + y z - z x = 0$ $y = z ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?$ $O bviously false!$
	Commented by HongKing last updated on 07/Nov/21

	$how my dear S er$
	Commented by mr W last updated on 07/Nov/21

	$y - z = 0 o r - x^{2} - x y + y z - z x = 0$ $y - z = 0 m u s t b e r e j e c t e d .$
	Commented by Rasheed.Sindhi last updated on 07/Nov/21

	$S i r, {d o e s n}^{'} t^{'} A o r B^{'} m e a n :$ $A o r B i s t r u e w h e n$ $a n y o n e o r b o t h a r e t r u e ?$
	Commented by mr W last updated on 07/Nov/21

	$b e c a u s e x^{2} (y - z) = 1 i s g i v e n .$
	Commented by Rasheed.Sindhi last updated on 07/Nov/21

	$G r a t e f u l l S i r!$
	Commented by mr W last updated on 07/Nov/21

	$A \times B = 0$ $m e a n s o n l y o n e f r o m A a n d B m u s t$ $b e z e r o . t h e o t h e r m a y b e z e r o, b u t$ ${m u s t n}^{'} t .$
	Commented by Rasheed.Sindhi last updated on 07/Nov/21

	$S i r, t h e n w h y i s y - z = 0 r e j e c t e d ?$
	Answered by mr W last updated on 06/Nov/21

	$\frac{y^{2} (z - x)}{x^{2} (y - z)} = - 1$ ${(\frac{y}{x})}^{2} \frac{\frac{z}{x} - 1}{\frac{y}{x} - \frac{z}{x}} = - 1$ $l e t p = \frac{y}{x}, q = \frac{z}{x}$ $p^{2} (q - 1) = - (p - q)$ $(p - 1) [(p + 1) q - p] = 0 . . . (i)$ $\frac{z^{2} (x - y)}{x^{2} (y - z)} = 2$ ${(\frac{z}{x})}^{2} \frac{1 - \frac{y}{x}}{\frac{y}{x} - \frac{z}{x}} = 2$ $q^{2} (1 - p) = 2 (p - q)$ $q (q + 2) = (2 + q^{2}) p$ $p = \frac{q (q + 2)}{2 + q^{2}} . . . (i i)$ $c a s e 1 :$ $f r o m (i) : p = 1$ $f r o m (i i) : q = 1$ $\Rightarrow n o s o l u t i o n$ $c a s e 2 :$ $f r o m (i) : (p + 1) q - p = 0 \Rightarrow p = \frac{q}{1 - q}$ $\frac{q (q + 2)}{2 + q^{2}} = \frac{q}{1 - q}$ $q \neq 0$ $\frac{q + 2}{2 + q^{2}} = \frac{1}{1 - q}$ $2 q^{2} + q = 0$ $2 q + 1 = 0$ $\Rightarrow q = - \frac{1}{2}$ $\Rightarrow p = \frac{- \frac{1}{2}}{1 + \frac{1}{2}} = - \frac{1}{3}$ $x^{2} (- \frac{x}{3} + \frac{x}{2}) = 1$ $\frac{x^{3}}{6} = 1$ $\Rightarrow x = \sqrt[3]{6}$ $\Rightarrow y = - \frac{\sqrt[3]{6}}{3}$ $\Rightarrow z = - \frac{\sqrt[3]{6}}{2}$
	Commented by HongKing last updated on 07/Nov/21

	$perfect my dear S er, thank you very much$
	Commented by Tawa11 last updated on 07/Nov/21

	$Great sir$
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