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Question Number 44783 by Necxx last updated on 04/Oct/18

$${\left(x^2\right)}^2+{\left(y^2\right)}^2=97.......1$$$${\left(x^2\right)}^3+{\left(y^2\right)}^3=793.......2$$$$solvethesimultaneousequation$$

Answered by ajfour last updated on 05/Oct/18

$$a^2+b^2=97$$$$a^3+b^3=793$$$$$$$$\Rightarrow {(a+b)}^2-2ab=97$$$$\mathrm{\&}{(a+b)}^3-3ab(a+b)=793$$$$leta+b=r\mathrm{\&}ab=s$$$$\Rightarrow r^2-2s=97\mathrm{\&}$$$$r^3-3rs=793$$$$\Rightarrow 2r^3-3r(r^3-97)=1586$$$$\Rightarrow r^3-291r+1586=0$$$$solutionofacubicoftheform$$$$x^3+px+q=0is$$$$\mathit{x}={(-\frac{\mathit{q}}{2}+\sqrt{\frac{{\mathit{q}}^2}{4}+\frac{{\mathit{p}}^3}{27}})}^{1/3}$$$$-{(\frac{\mathit{q}}{2}+\sqrt{\frac{{\mathit{q}}^2}{4}+\frac{{\mathit{p}}^3}{27}})}^{1/3}$$$$\left(ifithasonerealroot\right)$$$$Herewehave$$$${\mathit{r}}^3-291\mathit{r}+1586=0$$$$calculatorprovidestherealroot,$$$$\mathit{r}=13,\mathit{s}=\frac{r^2-97}{2}=\frac{169-97}{2}$$$$\Rightarrow \mathit{s}=36$$$$\Rightarrow a+b=13,ab=36$$$$x=\pm 9,y=\pm 4or$$$$x=\pm 4,y=\pm 9.$$

Commented by Necxx last updated on 05/Oct/18

$$pleasesolvehowr=13.Thanks$$

Commented by Necxx last updated on 05/Oct/18

$$Oh....Ireallydidntremember$$$$thecubicformsyouandMrMJS$$$$used.Thankssomuch$$

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