Determine the value of x+y if { ((x^3 +y^3 =1)),(((x+y)(x+1)(y+1)=2)) :}


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Question Number 98104 by bemath last updated on 11/Jun/20
$Determine the value of x+y if { ((x^3 +y^3 =1)),(((x+y)(x+1)(y+1)=2)) :}$
$Determine the value of x + y$ $if {\begin{cases} x^{3} + y^{3} = 1 \\ (x + y) (x + 1) (y + 1) = 2 \end{cases}$
Commented by bemath last updated on 12/Jun/20

$thank you all$
	Answered by mr W last updated on 11/Jun/20

	$u = x + y$ $v = x y$ $x^{3} + y^{3} = {(x + y)}^{3} - 3 x y (x + y) = 1$ $\Rightarrow u^{3} - 3 u v = 1$ $\Rightarrow v = \frac{u^{3} - 1}{3 u}$ $(x + y) (x y + x + y + 1) = 2$ $\Rightarrow u (u + v + 1) = 2$ $\Rightarrow u (u + \frac{u^{3} - 1}{3 u} + 1) = 2$ $\Rightarrow u^{3} + 3 u^{2} + 3 u - 1 = 6$ $\Rightarrow u^{3} + 3 u^{2} + 3 u + 1 = 8$ $\Rightarrow {(u + 1)}^{3} = 8 = 2^{3}$ $\Rightarrow u + 1 = 2, 2 ω, 2 ω^{2}$ $\Rightarrow u = x + y = 1 (r e a l)$ $\Rightarrow u = x + y = 2 ω - 1 = - 2 + i \sqrt{3}$ $\Rightarrow u = x + y = 2 ω^{2} - 1 = - 2 - i \sqrt{3}$
	Commented by Lekhraj last updated on 11/Jun/20

	${(u + 1)}^{3} = 8$
	Commented by mr W last updated on 11/Jun/20

	$y e s$
	Answered by Rasheed.Sindhi last updated on 11/Jun/20
	$if { ((x^3 +y^3 =1.......................(i))),(((x+y)(x+1)(y+1)=2.....(ii))) :} (ii)⇒ xy=(2/(x+y))−(x+y)−1.....(iii) (i)⇒ (x+y)^3 −3xy(x+y)=1......(iv) (iii)&(iv)⇒ (x+y)^3 −3((2/(x+y))−(x+y)−1)(x+y)=1 (x+y)^3 −3(2−(x+y)^2 −(x+y))=1 (x+y)^3 −6+3(x+y)^2 +3(x+y))=1 (x+y)^3 +3(x+y)^2 +3(x+y)−7=0 t^3 +3t^2 +3t−7=0 −7=(±1)(∓7) t=1 satisfy the equation x+y=1 ∴t−1 is factor of t^3 +3t^2 +3t−7 (((1)),1,3,3,(−7)),(,,1,4,( 7)),(,1,4,7,( 0)) ) t^2 +4t+7=0 t=((−4±(√(16−28)))/2) t=((−4±2i(√3))/2) x+y=((−4±2i(√3))/2)=−2±i(√3) x+y=1, −2±i(√3)$
	$if {\begin{cases} x^{3} + y^{3} = 1 . . . . . . . . . . . . . . . . . . . . . . . (i) \\ (x + y) (x + 1) (y + 1) = 2 . . . . . (ii) \end{cases}$ $(ii) \Rightarrow$ $xy = \frac{2}{x + y} - (x + y) - 1 . . . . . (iii)$ $(i) \Rightarrow$ ${(x + y)}^{3} - 3 xy (x + y) = 1 . . . . . . (iv)$ $(iii) & (iv) \Rightarrow$ ${(x + y)}^{3} - 3 (\frac{2}{x + y} - (x + y) - 1) (x + y) = 1$ ${(x + y)}^{3} - 3 (2 - {(x + y)}^{2} - (x + y)) = 1$ ${(x + y)}^{3} - 6 + 3 {(x + y)}^{2} + 3 (x + y)) = 1$ ${(x + y)}^{3} + 3 {(x + y)}^{2} + 3 (x + y) - 7 = 0$ $t^{3} + {3 t}^{2} + 3 t - 7 = 0$ $- 7 = (\pm 1) (\mp 7)$ $t = 1 satisfy the equation$ $x + y = 1$ $∴ t - 1 is factor of t^{3} + {3 t}^{2} + 3 t - 7$ $(\begin{matrix} 1) & 1 & 3 & 3 & - 7 \\ 1 & 4 & 7 \\ 1 & 4 & 7 & 0 \end{matrix})$ $t^{2} + 4 t + 7 = 0$ $t = \frac{- 4 \pm \sqrt{16 - 28}}{2}$ $t = \frac{- 4 \pm 2 i \sqrt{3}}{2}$ $x + y = \frac{- 4 \pm 2 i \sqrt{3}}{2} = - 2 \pm i \sqrt{3}$ $x + y = 1, - 2 \pm i \sqrt{3}$
	Answered by 1549442205 last updated on 11/Jun/20
	$(2)⇔(x+y)[xy+(x+y)+1]=2 (1)⇔(x+y)^3 −3xy(x+y)=1 Put x+y=u,xy=v we have { ((u^3 −3uv=1(3))),((u(u+v+1)=2(4))) :} From (4) we ger uv=2−u^2 −u.Replace into (3) we get u^3 −3(2−u^2 −u)=1⇔u^3 +3u^2 +3u−7=9 ⇔(u−1)(u^2 +4u+7)=0⇔u−1=0(as u^2 +4u+7=(u+2)^2 +3>0) ,so u=1,replace into (4) we get v=0.We have { ((x+y=1)),((xy=0)) :}⇔(x;y)∈{(0;1);(1;0)} Thus,our system of equations has two soluions:(x;y)∈{(0;1);(1;0)}$
	$(2) \Leftrightarrow (x + y) [xy + (x + y) + 1] = 2$ $(1) \Leftrightarrow {(x + y)}^{3} - 3 xy (x + y) = 1$ $Put x + y = u, xy = v we have {\begin{cases} u^{3} - 3 uv = 1 (3) \\ u (u + v + 1) = 2 (4) \end{cases}$ $From (4) we ger uv = 2 - u^{2} - u . Replace into (3) we get$ $u^{3} - 3 (2 - u^{2} - u) = 1 \Leftrightarrow u^{3} + {3 u}^{2} + 3 u - 7 = 9$ $\Leftrightarrow (u - 1) (u^{2} + 4 u + 7) = 0 \Leftrightarrow u - 1 = 0 (as u^{2} + 4 u + 7 = {(u + 2)}^{2} + 3 > 0)$ $, so u = 1, replace into (4) we get v = 0 . We have$ ${\begin{cases} x + y = 1 \\ xy = 0 \end{cases} \Leftrightarrow (x; y) \in {(0; 1); (1; 0)}$ $Thus, our system of equations has two$ $soluions : (x; y) \in {(0; 1); (1; 0)}$
	Answered by MJS last updated on 11/Jun/20
	$x=u−v∧y=u+v { ((2u^3 +6uv^2 −1=0)),((2(u^3 +2u^2 +u−1)−2uv^2 =0)) :} ⇒ v^2 =((1−2u^3 )/(6u))=((u^3 +2u^2 +u−1)/u) ⇒ u^3 +(3/2)u^2 +(3/4)u−(7/8)=0 u=t−(1/2) t^3 −1=0 t=1∨t=−(1/2)±((√3)/2)i u=(1/2)∨u=−1±((√3)/2)i x=u−v∧y=u+v ⇒ x+y=2u answer is 1∨−2±(√3)i$
	$x = u - v \land y = u + v$ ${\begin{cases} 2 u^{3} + 6 {u v}^{2} - 1 = 0 \\ 2 (u^{3} + 2 u^{2} + u - 1) - 2 {u v}^{2} = 0 \end{cases}$ $\Rightarrow$ $v^{2} = \frac{1 - 2 u^{3}}{6 u} = \frac{u^{3} + 2 u^{2} + u - 1}{u}$ $\Rightarrow$ $u^{3} + \frac{3}{2} u^{2} + \frac{3}{4} u - \frac{7}{8} = 0$ $u = t - \frac{1}{2}$ $t^{3} - 1 = 0$ $t = 1 \lor t = - \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$ $u = \frac{1}{2} \lor u = - 1 \pm \frac{\sqrt{3}}{2} i$ $x = u - v \land y = u + v \Rightarrow x + y = 2 u$ $answer is 1 \lor - 2 \pm \sqrt{3} i$
	Answered by maths mind last updated on 11/Jun/20
	$(x+y)(x+1)(y+1) =(x+y)(xy+x+y+1) ={xy(x+y)+(x+y)^2 +(x+y)=2}×3..E x^3 +y^3 =1..G E+G=(x+y)^3 +3(x+y)^2 +3(x+y)=7 ⇔(x+y)^3 +3(x+y)^2 +3(x+y)+1=8 ⇔(1+x+y)^3 =8⇒1+x+y=2e^(i((2kπ)/3)) ⇒x+y=2e^(i((kπ)/3)) −1,k∈{0,1,2}$
	$(x + y) (x + 1) (y + 1)$ $= (x + y) (x y + x + y + 1)$ $= {x y (x + y) + {(x + y)}^{2} + (x + y) = 2} \times 3 . . E$ $x^{3} + y^{3} = 1 . . G$ $E + G = {(x + y)}^{3} + 3 {(x + y)}^{2} + 3 (x + y) = 7$ $\Leftrightarrow {(x + y)}^{3} + 3 {(x + y)}^{2} + 3 (x + y) + 1 = 8$ $\Leftrightarrow {(1 + x + y)}^{3} = 8 \Rightarrow 1 + x + y = 2 e^{i \frac{2 k π}{3}}$ $\Rightarrow x + y = 2 e^{i \frac{k π}{3}} - 1, k \in {0, 1, 2}$
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