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Question Number 61283 by naka3546 last updated on 01/Jun/19

$$(x+y)(x^2+y^2)(x^3+y^3)=2$$$$(x^4+y^4)(x^6+y^6)(x^8+y^8)=4$$$$(x^3+y^3)(x^5+y^5)(x^7+y^7)=6$$$$(x^4+y^4)(x^5+y^5)(x^9+y^9)(x^{10}+y^{10})=?$$

Answered by MJS last updated on 02/Jun/19

$$\mathrm{let}$$$$x=p-q,y=p+q\mathrm{with}p,q\in \mathbb{C}$$$$\Rightarrow \mathrm{the}{1}^{\mathrm{st}}\mathrm{equation}\mathrm{then}\mathrm{transforms}\mathrm{to}$$$$q^4+\frac{4p^2}{3}{q}^2+\frac{4p^6-1}{12p^2}=0$$$$\Rightarrow q=\pm \sqrt{-\frac{2p^2}{3}\pm \frac{\sqrt{4p^6+3}}{6p}}\left[4\mathrm{solutions}\right]$$$$\mathrm{approximately}\mathrm{solving}\mathrm{the}{3}^{\mathrm{rd}}\mathrm{equation}$$$$\mathrm{finally}\mathrm{gives}\mathrm{the}\mathrm{following}\mathrm{pairs}\left(\begin{array}{c}x\\ y\end{array}\right):$$$$\left(\begin{array}{c}1.126879\\ -.02685083\end{array}\right)\left(\begin{array}{c}1.008297+.4838797\mathrm{i}\\ 1.008297-.4838797\mathrm{i}\end{array}\right)$$$$\left(\begin{array}{c}1.116944-.009290636\mathrm{i}\\ -.3248961-.8386042\mathrm{i}\end{array}\right)\left(\begin{array}{c}1.116944+.009290636\mathrm{i}\\ -.3248961+.8386042\mathrm{i}\end{array}\right)$$$$\left[\mathrm{of}\mathrm{course}\mathrm{we}\mathrm{can}\mathrm{exchange}x\mathrm{with}y\right]$$$$\mathrm{none}\mathrm{of}\mathrm{these}\mathrm{values}\mathrm{satisfy}\mathrm{the}{2}^{\mathrm{nd}}\mathrm{equation}$$$$$$$$\Rightarrow \mathrm{no}\mathrm{solution}\mathrm{in}\mathbb{C}$$

Commented by alphaprime last updated on 02/Jun/19

Yes sir absolutely correct said , it got no solutions

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