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Question Number 201722 by Calculusboy last updated on 11/Dec/23

Commented by Frix last updated on 11/Dec/23

$There also should be complex solutions$
	Answered by Sutrisno last updated on 11/Dec/23

	$x^{2} + y^{2} + x y = 3$ ${(x + y)}^{2} - x y = 3$ $x y = {(x + y)}^{2} - 3 . . . (1)$ $((x + y) + 1) (4 + {(x + y)}^{2} - 3 + 2 (x + y)) = 27$ $((x + y) + 1) ({(x + y)}^{2} + 2 (x + y) + 1) = 27$ $((x + y) + 1) {((x + y) + 1)}^{2} = 27$ ${((x + y) + 1)}^{3} = 27$ $x + y = 2 . . . (2)$ $x y = 2^{2} - 3 \to x y = 1$ $x (2 - x) = 1$ $x^{2} - 2 x + 1 = 0$ $x = 1, y = 1$
	Commented by Calculusboy last updated on 11/Dec/23

	$t h a n k s s i r$
	Answered by esmaeil last updated on 12/Dec/23

	$x + y = s \to S^{2} - p = 3$ $x y = p \to (s + 1) (4 + p + 2 s) = 27 \to$ $(s + 1) ((s^{2} + 1 + 2 s) = 27 \to$ $(s + 1) = 3 \to s = 2 \to p = 1$ $a^{2} - 2 a + 1 = 0 \to x = y = 1$
	Answered by Rasheed.Sindhi last updated on 12/Dec/23
	$Slightly different way { ((x^2 +y^2 +xy=3.......(i))),(((x+y+1)(4+xy+2x+2y)=27...(ii))) :} (ii)⇒(x+y+1)(1+3+xy+2x+2y)=27 (x+y+1)(1+x^2 +y^2 +xy+xy+2x+2y)=27 (x+y+1)(1+x^2 +y^2 +2xy+2x+2y)=27 (x+y+1)( (x+y)^2 +2(x+y)+1 )=27 (x+y+1) (x+y+1)^2 =27 (x+y+1)^3 =27 x+y+1=3 x+y=2.....(iii) x^2 +y^2 +2xy=4 x^2 +y^2 +xy+xy=4 3+xy=4 xy=1.......(i) (iii) & (iv): x=1,y=1$
	$S l i g h t l y d i f f e r e n t w a y$ ${\begin{cases} x^{2} + y^{2} + x y = 3 . . . . . . . (i) \\ (x + y + 1) (4 + x y + 2 x + 2 y) = 27 . . . (i i) \end{cases}$ $(i i) \Rightarrow (x + y + 1) (1 + 3 + x y + 2 x + 2 y) = 27$ $(x + y + 1) (1 + x^{2} + y^{2} + x y + x y + 2 x + 2 y) = 27$ $(x + y + 1) (1 + x^{2} + y^{2} + 2 x y + 2 x + 2 y) = 27$ $(x + y + 1) ({(x + y)}^{2} + 2 (x + y) + 1) = 27$ $(x + y + 1) {(x + y + 1)}^{2} = 27$ ${(x + y + 1)}^{3} = 27$ $x + y + 1 = 3$ $x + y = 2 . . . . . (i i i)$ $x^{2} + y^{2} + 2 x y = 4$ $x^{2} + y^{2} + x y + x y = 4$ $3 + x y = 4$ $x y = 1 . . . . . . . (i)$ $(i i i) & (i v) : x = 1, y = 1$
	Commented by Calculusboy last updated on 15/Dec/23

	$t h a n k s s i r$
	Answered by Rasheed.Sindhi last updated on 12/Dec/23
	$A way leading determinant (((ALL))) the solutions: { ((x^2 +y^2 +xy=3....(i))),(((x+y+1)(4+xy+2x+2y)=27...(ii))) :} (ii)⇒(x+y+1)(1+3+xy+2x+2y)=(3)^3 (x+y+1)(1+x^2 +y^2 +xy+xy+2x+2y)=(3)^3 (x+y+1)( (x+y)^2 +2(x+y)+1 )=(3)^3 (x+y+1)(x+y+1)^2 =(3)^3 (x+y+1)^3 =(3)^(3 ) (x+y+1)^3 −(3)^(3 ) =0 {(x+y+1)−3}{(x+y+1)^2 +3(x+y+1)+9}=0^(★★) x+y+1−3=0_(→real roots) ∣ (x+y+1)^2 +3(x+y+1)+9=0^★ _(→complex roots) x+y=2⇒y=2−x (i)⇒x^2 +(2−x)^2 +x(2−x)=3 x^2 −2x+1=0 (x−1)^2 =0 x=1^✓ ⇒y=2−x=2−1=1^✓ ^★ x+y+1=((−3±(√(9−4(9)(1))))/2)=((−3±3(√3))/2) x+y=((−3±3(√3))/2)−1=((−5±3(√3))/2) y=((−5±3(√3))/2)−x (i)⇒(x+y)^2 −xy=3⇒(((−5±3(√3))/2))^2 −x(((−5±3(√3))/2)−x)=3 let ((−5±3(√3))/2)=a a^2 −x(a−x)−3=0 x^2 −ax+a^2 −3=0 x=((a±(√(a^2 −4(a^2 −3))))/2)=((a±(√(12−3a^2 )))/2) y=a−x=a−((a±(√(12−3a^2 )))/2)=((a∓(√(12−3a^2 )))/2) where a= ((−5±3(√3))/2)$
	$A way leading \begin{matrix} ALL \end{matrix} the solutions :$ ${\begin{cases} x^{2} + y^{2} + x y = 3 . . . . (i) \\ (x + y + 1) (4 + x y + 2 x + 2 y) = 27 . . . (i i) \end{cases}$ $(i i) \Rightarrow (x + y + 1) (1 + 3 + x y + 2 x + 2 y) = {(3)}^{3}$ $(x + y + 1) (1 + x^{2} + y^{2} + x y + x y + 2 x + 2 y) = {(3)}^{3}$ $(x + y + 1) ({(x + y)}^{2} + 2 (x + y) + 1) = {(3)}^{3}$ $(x + y + 1) {(x + y + 1)}^{2} = {(3)}^{3}$ ${(x + y + 1)}^{3} = {(3)}^{3}$ ${(x + y + 1)}^{3} - {(3)}^{3} = 0$ ${(x + y + 1) - 3} {{(x + y + 1)}^{2} + 3 (x + y + 1) + 9} = 0^{★ ★}$ $Missing \left or extra \right$ $x + y = 2 \Rightarrow y = 2 - x$ $(i) \Rightarrow x^{2} + {(2 - x)}^{2} + x (2 - x) = 3$ $x^{2} - 2 x + 1 = 0$ ${(x - 1)}^{2} = 0$ $x = 1^{✓} \Rightarrow y = 2 - x = 2 - 1 = 1^{✓}$ $^{★} x + y + 1 = \frac{- 3 \pm \sqrt{9 - 4 (9) (1)}}{2} = \frac{- 3 \pm 3 \sqrt{3}}{2}$ $x + y = \frac{- 3 \pm 3 \sqrt{3}}{2} - 1 = \frac{- 5 \pm 3 \sqrt{3}}{2}$ $y = \frac{- 5 \pm 3 \sqrt{3}}{2} - x$ $(i) \Rightarrow {(x + y)}^{2} - x y = 3 \Rightarrow {(\frac{- 5 \pm 3 \sqrt{3}}{2})}^{2} - x (\frac{- 5 \pm 3 \sqrt{3}}{2} - x) = 3$ $l e t \frac{- 5 \pm 3 \sqrt{3}}{2} = a$ $a^{2} - x (a - x) - 3 = 0$ $x^{2} - a x + a^{2} - 3 = 0$ $x = \frac{a \pm \sqrt{a^{2} - 4 (a^{2} - 3)}}{2} = \frac{a \pm \sqrt{12 - 3 a^{2}}}{2}$ $y = a - x = a - \frac{a \pm \sqrt{12 - 3 a^{2}}}{2} = \frac{a \mp \sqrt{12 - 3 a^{2}}}{2}$ $w h e r e a = \frac{- 5 \pm 3 \sqrt{3}}{2}$
	Commented by Rasheed.Sindhi last updated on 12/Dec/23

	$^{★ ★} x + y + 1 - 3 = 0 ∣ {(x + y + 1)}^{2} + 3 (x + y + 1) + 9 = 0$ $x + y + 1 = 3, 3 ω, 3 ω^{2}$ $x + y = 2, 3 ω - 1, 3 ω^{2} - 1$ $(i) \Rightarrow {(x + y)}^{2} - x y = 3 \Rightarrow x y = {(x + y)}^{2} - 3$ $x y = 4 - 3, {(3 ω - 1)}^{2} - 3, {(3 ω^{2} - 1)}^{2} - 3$ $. . . . . .$ $. . . .$
	Commented by Calculusboy last updated on 15/Dec/23

	$n i c e s o l u t i o n s i r$
	Answered by ajfour last updated on 12/Dec/23

	$x + y = s, x y = m$ $s^{2} = 3 + m$ $(s + 1) (2 s + m + 4) = 27$ $(s + 1) (s^{2} + 2 s + 1) = 27$ $s = - 1 + 3, - 1 + 3 ω, - 1 + 3 ω^{2}$ $x, y = \frac{s \pm \sqrt{3 (4 - s^{2})}}{2}$ $s a y i f s = 2$ $x, y = \frac{s}{2} = 1$
	Commented by Calculusboy last updated on 15/Dec/23

	$t h a n k s s i r$
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