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Question Number 175373 by infinityaction last updated on 28/Aug/22

$$\mathrm{if}{x}^6+y^6=9$$$$x^4+y^4=5$$$$\mathrm{then}{x}^2+y^2=?$$

Answered by mr W last updated on 28/Aug/22

$$say{x}^2+y^2=a$$$$x^4+y^4+2x^2{y}^2=a^2$$$$\Rightarrow x^2{y}^2=\frac{a^2-5}{2}$$$$(x^4+y^4)(x^2+y^2)=x^6+y^6+x^2{y}^2(x^2+y^2)$$$$5a=9+\frac{(a^2-5)a}{2}$$$$a^3-15a+18=0$$$$(a-3)(a^2+3a-6)=0$$$$\Rightarrow a=3,\frac{-3\pm \sqrt{33}}{2}$$

Commented by Tawa11 last updated on 28/Aug/22

$$\mathrm{Great}\mathrm{sirs}$$

Answered by Rasheed.Sindhi last updated on 28/Aug/22

$$x^4+y^4={(x^2+y^2)}^2-2x^2{y}^2=5$$$$x^2{y}^2=\frac{{(x^2+y^2)}^2-5}{2}$$$$x^6+y^6={(x^2+y^2)}^3-3x^2{y}^2(x^2+y^2)=9$$$${(x^2+y^2)}^3-3\left(\frac{{(x^2+y^2)}^2-5}{2}\right)(x^2+y^2)=9$$$$x^2+y^2=p$$$$p^3-3\left(\frac{p^2-5}{2}\right)\left(p\right)=9$$$$2p^3-3p^3+15p-18=0$$$$p^3-15p+18=0$$$$p=x^2+y^2=3,\frac{-3\pm \sqrt{33}}{2}$$

Answered by Rasheed.Sindhi last updated on 28/Aug/22

$$▸x^6+y^6=(x^2+y^2)(x^4+y^4-x^2{y}^2)=9$$$$(x^2+y^2)(5-x^2{y}^2)=9$$$$x^2{y}^2=5-\frac{9}{x^2+y^2}=\frac{5(x^2+y^2)-9}{x^2+y^2}$$$$▸x^6+y^6={(x^2+y^2)}^3-3x^2{y}^2(x^2+y^2)=9$$$${(x^2+y^2)}^3-3\left(\frac{5(x^2+y^2)-9}{\cancel{x^2+y^2}}\right)\cancel{(x^2+y^2)}=9$$$$x^2+y^2=p$$$$p^3-3(5p-9)=9$$$$p^3-15p+18=0$$$$p=x^2+y^2=3,-\frac{3}{2}\pm \frac{\sqrt{33}}{2}$$$$$$

Commented by infinityaction last updated on 28/Aug/22

$$thanks$$

Answered by ajfour last updated on 29/Aug/22

$${(x^2+y^2)}^3=p^3=9+3x^2{y}^{2}p$$$$p^2=5+2x^2{y}^2$$$$\Rightarrow 2p^3=18+3p(p^2-5)$$$$\Rightarrow p^3-15p+18=0$$$$p=3,\alpha ,\beta$$

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