Question

Solve the following equations:
$x^4+x^2{y}^2+y^4=931$,
$x^2-xy+y^2=19$.

Answer 1

$x^2-xy+y^2=19x^2+y^2=19+xy......\left(i\right)x^4+y^2{x}^2+y^4=931$

Using ${(a+b)}^2{=a}^2+b^2+2ab$

$(x^2+y^2{)}^2-x^2{y}^2=\mathrm{931.....}\left(ii\right)$

Substituting $\left(i\right)$ in $\left(ii\right)$

${(19+xy)}^2-x^2{y}^2=931361+x^2{y}^2+38xy-x^2{y}^2=93138xy=570\Rightarrow xy=15y=\frac{15}{x}{x}^2+y^2=19+xy{x}^2+{\left(\frac{15}{x}\right)}^2=19+15x^4+225=24x^2{x}^4-34x^2+225=0x^4-9x^2-25x^2+225=0x^2(x^2-9)-25(x^2-9)=0(x^2-25)(x^2-9)=0x^2=25,9x=\sqrt{25}\sqrt{9}x=\pm 5,\pm 3$

Now, $y=\frac{15}{x}$

$x=\pm 5\Rightarrow y=\frac{15}{\pm 5}=\pm 3x\pm 3\Rightarrow y=\frac{15}{\pm 3}=\pm 5$