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Question Number 164787 by cortano1 last updated on 22/Jan/22

	Answered by mr W last updated on 22/Jan/22
	$xy[(x+y)^2 −2xy]=((10(x+y)^2 )/9) xy[(x+y)^3 −3xy(x+y)]=((2(x+y)^3 )/3) let u=x+y, v=xy { ((v(u^2 −2v)=((10u^2 )/9) ...(i))),((vu(u^2 −3v)=((2u^3 )/3) ...(ii))) :} from (ii): u=0 ⇒v=0 ⇒x=y=0 or u≠0 ⇒v≠0 v(u^2 −3v)=((2u^2 )/3) ...(iii) (i)/(ii): ((u^2 −2v)/(u^2 −3v))=(5/3) 2u^2 =9v ⇒v=((2u^2 )/9) put this into (iii): ((2u^2 )/9)(u^2 −3×((2u^2 )/9))=((2u^2 )/3) u^2 =9 ⇒u=±3 ⇒v=2 x,y are roots of z^2 −uz+v=0 x,y=((u±(√(u^2 −4v)))/2)=((±3±1)/2)=2, 1 or −2,−1 summary: (x,y)=(0,0), (−2,−1),(−1,−2),(2,1),(1,2)$
	$x y [{(x + y)}^{2} - 2 x y] = \frac{10 {(x + y)}^{2}}{9}$ $x y [{(x + y)}^{3} - 3 x y (x + y)] = \frac{2 {(x + y)}^{3}}{3}$ $l e t u = x + y, v = x y$ ${\begin{cases} v (u^{2} - 2 v) = \frac{10 u^{2}}{9} . . . (i) \\ v u (u^{2} - 3 v) = \frac{2 u^{3}}{3} . . . (i i) \end{cases}$ $f r o m (i i) :$ $u = 0$ $\Rightarrow v = 0$ $\Rightarrow x = y = 0$ $o r$ $u \neq 0 \Rightarrow v \neq 0$ $v (u^{2} - 3 v) = \frac{2 u^{2}}{3} . . . (i i i)$ $(i) / (i i) :$ $\frac{u^{2} - 2 v}{u^{2} - 3 v} = \frac{5}{3}$ $2 u^{2} = 9 v$ $\Rightarrow v = \frac{2 u^{2}}{9}$ $p u t t h i s i n t o (i i i) :$ $\frac{2 u^{2}}{9} (u^{2} - 3 \times \frac{2 u^{2}}{9}) = \frac{2 u^{2}}{3}$ $u^{2} = 9$ $\Rightarrow u = \pm 3$ $\Rightarrow v = 2$ $x, y a r e r o o t s o f z^{2} - u z + v = 0$ $x, y = \frac{u \pm \sqrt{u^{2} - 4 v}}{2} = \frac{\pm 3 \pm 1}{2} = 2, 1 o r - 2, - 1$ $s u m m a r y :$ $(x, y) = (0, 0), (- 2, - 1), (- 1, - 2), (2, 1), (1, 2)$
	Commented by leonhard77 last updated on 22/Jan/22

	$(i) : (i i) \Rightarrow \frac{5}{3 u}$
	Commented by mr W last updated on 22/Jan/22

	$y o u s h o u l d l o o k a t b o t h s i d e s!$
	Commented by cortano1 last updated on 22/Jan/22

	Commented by Tawa11 last updated on 22/Jan/22

	$Great sirs$
	Answered by MJS_new last updated on 22/Jan/22
	$let y=px { ((p(p^2 +1)x^4 =((10)/9)(p+1)^2 x^2 ⇒ x=y=0)),((p(p+1)(p^2 −p+1)x^5 =(2/3)(p+1)^3 x^3 ⇒ x=y=0 ∨ p=−1)) :} { ((p(p^2 +1)x^2 =((10)/9)(p+1)^2 )),((p(p^2 −p+1)x^2 =(2/3)(p+1)^2 )) :} { ((x^2 =((10(p+1)^2 )/(9p(p^2 +1))))),((x^2 =((2(p+1)^2 )/(3p(p^2 −p+1))))) :} ⇒ ((10(p+1)^2 )/(9p(p^2 +1)))=((2(p+1)^2 )/(3p(p^2 −p+1))) ⇒ p=−1∨p=(1/2)∨p=2 the rest is easy$
	$let y = p x$ ${\begin{cases} p (p^{2} + 1) x^{4} = \frac{10}{9} {(p + 1)}^{2} x^{2} \Rightarrow x = y = 0 \\ p (p + 1) (p^{2} - p + 1) x^{5} = \frac{2}{3} {(p + 1)}^{3} x^{3} \Rightarrow x = y = 0 \lor p = - 1 \end{cases}$ ${\begin{cases} p (p^{2} + 1) x^{2} = \frac{10}{9} {(p + 1)}^{2} \\ p (p^{2} - p + 1) x^{2} = \frac{2}{3} {(p + 1)}^{2} \end{cases}$ ${\begin{cases} x^{2} = \frac{10 {(p + 1)}^{2}}{9 p (p^{2} + 1)} \\ x^{2} = \frac{2 {(p + 1)}^{2}}{3 p (p^{2} - p + 1)} \end{cases}$ $\Rightarrow \frac{10 {(p + 1)}^{2}}{9 p (p^{2} + 1)} = \frac{2 {(p + 1)}^{2}}{3 p (p^{2} - p + 1)} \Rightarrow p = - 1 \lor p = \frac{1}{2} \lor p = 2$ $the rest is easy$
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