{ ((x^3 +y^3 =3xy)),((x^4 +y^4 =4xy)) :} [x,y≠0]


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Question Number 62281 by behi83417@gmail.com last updated on 19/Jun/19
${ ((x^3 +y^3 =3xy)),((x^4 +y^4 =4xy)) :} [x,y≠0]$
${\begin{cases} x^{3} + y^{3} = 3 xy \\ x^{4} + y^{4} = 4 xy \end{cases} [x, y \neq 0]$
	Answered by mr W last updated on 19/Jun/19
	$let u=x+y, v=xy x^3 +y^3 =3xy (x+y)^3 =x^3 +y^3 +3xy(x+y) (x+y)^3 =3xy+3xy(x+y) u^3 =3v(1+u) ⇒v=(u^3 /(3(1+u))) x^4 +y^4 =(x^2 +y^2 )^2 −2x^2 y^2 =[(x+y)^2 −2xy]^2 −2(xy)^2 =4xy ⇒(u^2 −2v)^2 −2v^2 =4v ⇒[u^2 −((2u^3 )/(3(1+u)))]^2 −2×(u^6 /(9(1+u)^2 ))=4×(u^3 /(3(1+u))) ⇒[3u^2 (1+u)−2u^3 ]^2 −2u^6 =12u^3 (1+u) ⇒u^4 (3+u)^2 −2u^6 =12u^3 (1+u) ⇒u=0⇒x=y=0 ⇒u(3+u)^2 −2u^3 =12(1+u) ⇒u^3 −6u^2 +3u+12=0 let u=s+2 ⇒s^3 −9s+2=0 Δ=(−3)^3 +1^2 <0 ⇒3 real roots ⇒s=2(√3) sin ((1/3) sin^(−1) (1/(3(√3)))+((2kπ)/3)) ⇒u=2+2(√3) sin ((1/3) sin^(−1) (1/(3(√3)))+((2kπ)/3)) ⇒v=(u^3 /(3(1+u))) ⇒(x−y)^2 =(x+y)^2 −4xy=u^2 −4v ⇒x−y=±(√(u^2 −4v)) ⇒x+y=u ⇒x=((u±(√(u^2 −4v)))/2) ⇒y=((u∓(√(u^2 −4v)))/2) { ((x=1.426766 / 0.796696 / 2.44101+0.79718i / 2.44101−0.79718i / −0.55274+1.99092i /−0.55274−1.99092i )),((y=0.796696 / 1.426766 / 2.44101−0.79718i / 2.44101+0.79718i / −0.55274−1.99092i / −0.55274+1.99092i )) :}$
	$l e t u = x + y, v = x y$ $x^{3} + y^{3} = 3 x y$ ${(x + y)}^{3} = x^{3} + y^{3} + 3 x y (x + y)$ ${(x + y)}^{3} = 3 x y + 3 x y (x + y)$ $u^{3} = 3 v (1 + u)$ $\Rightarrow v = \frac{u^{3}}{3 (1 + u)}$ $x^{4} + y^{4} = {(x^{2} + y^{2})}^{2} - 2 x^{2} y^{2} = {[{(x + y)}^{2} - 2 x y]}^{2} - 2 {(x y)}^{2} = 4 x y$ $\Rightarrow {(u^{2} - 2 v)}^{2} - 2 v^{2} = 4 v$ $\Rightarrow {[u^{2} - \frac{2 u^{3}}{3 (1 + u)}]}^{2} - 2 \times \frac{u^{6}}{9 {(1 + u)}^{2}} = 4 \times \frac{u^{3}}{3 (1 + u)}$ $\Rightarrow {[3 u^{2} (1 + u) - 2 u^{3}]}^{2} - 2 u^{6} = 12 u^{3} (1 + u)$ $\Rightarrow u^{4} {(3 + u)}^{2} - 2 u^{6} = 12 u^{3} (1 + u)$ $\Rightarrow u = 0 \Rightarrow x = y = 0$ $\Rightarrow u {(3 + u)}^{2} - 2 u^{3} = 12 (1 + u)$ $\Rightarrow u^{3} - 6 u^{2} + 3 u + 12 = 0$ $l e t u = s + 2$ $\Rightarrow s^{3} - 9 s + 2 = 0$ $Δ = {(- 3)}^{3} + 1^{2} < 0 \Rightarrow 3 r e a l r o o t s$ $\Rightarrow s = 2 \sqrt{3} \sin (\frac{1}{3} \sin^{- 1} \frac{1}{3 \sqrt{3}} + \frac{2 k π}{3})$ $\Rightarrow u = 2 + 2 \sqrt{3} \sin (\frac{1}{3} \sin^{- 1} \frac{1}{3 \sqrt{3}} + \frac{2 k π}{3})$ $\Rightarrow v = \frac{u^{3}}{3 (1 + u)}$ $\Rightarrow {(x - y)}^{2} = {(x + y)}^{2} - 4 x y = u^{2} - 4 v$ $\Rightarrow x - y = \pm \sqrt{u^{2} - 4 v}$ $\Rightarrow x + y = u$ $\Rightarrow x = \frac{u \pm \sqrt{u^{2} - 4 v}}{2}$ $\Rightarrow y = \frac{u \mp \sqrt{u^{2} - 4 v}}{2}$ ${\begin{cases} x = 1.426766 / 0.796696 / 2.44101 + 0.79718 i / 2.44101 - 0.79718 i / - 0.55274 + 1.99092 i / - 0.55274 - 1.99092 i \\ y = 0.796696 / 1.426766 / 2.44101 - 0.79718 i / 2.44101 + 0.79718 i / - 0.55274 - 1.99092 i / - 0.55274 + 1.99092 i \end{cases}$
	Commented by behi83417@gmail.com last updated on 19/Jun/19

	$thanks in advance dear master .$ $nice and smart solution .$
	Commented by behi83417@gmail.com last updated on 19/Jun/19

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