{ ((x^2 + 3xy + y^2 = −1)),((x^3 + y^3 = 7)) :}


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Question Number 112287 by weltr last updated on 07/Sep/20
${ ((x^2 + 3xy + y^2 = −1)),((x^3 + y^3 = 7)) :}$
${\begin{cases} x^{2} + 3 x y + y^{2} = - 1 \\ x^{3} + y^{3} = 7 \end{cases}$
	Answered by ajfour last updated on 07/Sep/20

	$x^{2} + y^{2} - x y = - 1 - 4 x y$ $x^{3} + y^{3} = - (x + y) (1 + 4 x y) = 7$ $& {(x + y)}^{2} = - 1 - x y$ $s a y x + y = s$ $\Rightarrow s (1 - 4 - 4 s^{2}) + 7 = 0$ $\Rightarrow (4 s^{2} + 3) s - 7 = 0$ $s = 1, - \frac{1}{2} \pm \frac{i \sqrt{6}}{2}$ $f o r r e a l x, y w e c o n s i d e r j u s t s = 1$ $\Rightarrow x + y = 1, x y = - 2$ $x, y = \frac{1}{2} \pm \sqrt{\frac{1}{4} + 2}$ $(x, y) \equiv (2, - 1) o r (- 1, 2)$
	Answered by MJS_new last updated on 07/Sep/20
	$let x=u−(√v)∧y=u+(√v) { ((5u^2 −v=−1 ⇒ v=5u^2 +1)),((2u^3 +6uv=7 ⇒ v=−((2u^3 −7)/(6u)))) :} 5u^2 +1=−((2u^3 −7)/(6u)) u^3 +(3/(16))u−(7/(32))=0 (u−(1/2))(u^2 +(1/2)u+(7/(16)))=0 u_1 =(1/2) u_(2, 3) =−(1/4)±((√6)/4)i v_1 =(9/4) v_(2, 3) =−(9/(16))∓((5(√6))/8)i x_1 =−1 y_1 =2 x_(2, 3) =−((2+(√(−18+2(√(681)))))/8)±((2(√6)+(√(18+2(√(681)))))/8)i y_(2, 3) =−((2−(√(−18+2(√(681)))))/8)±((2(√6)−(√(18+2(√(681)))))/8)i and of course we can exchange x with y due to symmetry$
	$let x = u - \sqrt{v} \land y = u + \sqrt{v}$ ${\begin{cases} 5 u^{2} - v = - 1 \Rightarrow v = 5 u^{2} + 1 \\ 2 u^{3} + 6 u v = 7 \Rightarrow v = - \frac{2 u^{3} - 7}{6 u} \end{cases}$ $5 u^{2} + 1 = - \frac{2 u^{3} - 7}{6 u}$ $u^{3} + \frac{3}{16} u - \frac{7}{32} = 0$ $(u - \frac{1}{2}) (u^{2} + \frac{1}{2} u + \frac{7}{16}) = 0$ $u_{1} = \frac{1}{2} u_{2, 3} = - \frac{1}{4} \pm \frac{\sqrt{6}}{4} i$ $v_{1} = \frac{9}{4} v_{2, 3} = - \frac{9}{16} \mp \frac{5 \sqrt{6}}{8} i$ $x_{1} = - 1 y_{1} = 2$ $x_{2, 3} = - \frac{2 + \sqrt{- 18 + 2 \sqrt{681}}}{8} \pm \frac{2 \sqrt{6} + \sqrt{18 + 2 \sqrt{681}}}{8} i$ $y_{2, 3} = - \frac{2 - \sqrt{- 18 + 2 \sqrt{681}}}{8} \pm \frac{2 \sqrt{6} - \sqrt{18 + 2 \sqrt{681}}}{8} i$ $and of course we can exchange x with y due$ $to symmetry$
	Answered by bemath last updated on 07/Sep/20
	$(x+y)^3 −3xy(x+y) = 7 (x+y)((x+y)^2 −3xy)= 7...(i) (x+y)^2 +xy = −1...(ii) let → { ((x+y = u)),((xy = v)) :} { ((u^3 −3uv = 7)),((u^2 +v = −1⇒v=−1−u^2 )) :} u^3 −3u(−1−u^2 )−7=0 u^3 +3u+3u^3 −7=0 4u^3 +3u−7 = 0 (u−1)(4u^2 +4u+7 ) = 0 u = 1 , u = ((−4 ± (√(16−112)))/8)$
	${(x + y)}^{3} - 3 xy (x + y) = 7$ $(x + y) ({(x + y)}^{2} - 3 xy) = 7 . . . (i)$ ${(x + y)}^{2} + xy = - 1 . . . (ii)$ $let \to {\begin{cases} x + y = u \\ xy = v \end{cases}$ ${\begin{cases} u^{3} - 3 uv = 7 \\ u^{2} + v = - 1 \Rightarrow v = - 1 - u^{2} \end{cases}$ $u^{3} - 3 u (- 1 - u^{2}) - 7 = 0$ $u^{3} + 3 u + {3 u}^{3} - 7 = 0$ ${4 u}^{3} + 3 u - 7 = 0$ $(u - 1) ({4 u}^{2} + 4 u + 7) = 0$ $u = 1, u = \frac{- 4 \pm \sqrt{16 - 112}}{8}$
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