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Question Number 12682 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 28/Apr/17

Answered by mrW1 last updated on 29/Apr/17

$$letu=x+y$$$$letv=xy$$$$$$$${(x+y)}^2=x^2+y^2+2xy=4+2xy$$$$\Rightarrow u^2=4+2v...\left(i\right)$$$$$$$$x^3+y^3=(x+y)(x^2+y^2-xy)=8$$$$(x+y)(4-xy)=8$$$$\Rightarrow u(4-v)=8...\left(ii\right)$$$$$$$$from\left(i\right):$$$$v=\frac{u^2}{2}-2$$$$into\left(ii\right):$$$$u(4-\frac{u^2}{2}+2)=8$$$$u(12-u^2)=16$$$$u^3-12u+16=0$$$$(u-2)(u-2)(u+4)=0$$$$\Rightarrow u=2or-4$$$$\Rightarrow v=0or6$$$$$$$$\Rightarrow x+y=2$$$$\Rightarrow xy=0$$$$\Rightarrow x^2+y^2-2xy=4$$$$\Rightarrow {(x-y)}^2=4$$$$\Rightarrow x-y=\pm 2$$$$\Rightarrow x=2or0$$$$\Rightarrow y=0or2$$$$$$$$\Rightarrow x+y=-4$$$$\Rightarrow xy=6$$$$\Rightarrow x^2+y^2-2xy=4-12=-8$$$$\Rightarrow {(x-y)}^2=-8$$$$\Rightarrow x-y=\pm 2\sqrt{2}i$$$$\Rightarrow x=-2\pm \sqrt{2}i$$$$\Rightarrow y=-2\mp \sqrt{2}i$$$$$$$$sothesolutionsare:$$$$\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}2\\ 0\end{array}\right)or\left(\begin{array}{c}0\\ 2\end{array}\right)or\left(\begin{array}{c}-2+\sqrt{2}i\\ -2-\sqrt{2}i\end{array}\right)or\left(\begin{array}{c}-2-\sqrt{2}i\\ -2+\sqrt{2}i\end{array}\right)$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 29/Apr/17

$$thanksalotdearmrW1.youareno.1.$$

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