[วิหก]

$x^{6}+y^{6}-(x^{2}+y^{2})(x^{4}+x^{2}y^{2})=0$
$x^{6}+y^{6}-x^{2}(x^{2}+y^{2})(x^{2}+y^{2})=0$
$x^{6}+y^{6}-x^{2}[(x^{2}+y^{2})^{2}]=0$
$x^{6}+y^{6}-x^{2}(x^{4}+2x^{4}y^{2}+x^{2}y^{4})=0$
$x^{6}+y^{6}-(x^{6}+2x^{4}y^{2}+x^{2}y^{4})=0$
$y^{6}-2x^{4}y^{2}-x^{2}y^{4}=0$
$y^{2}(y^4-2x^{4}-x^{2}y^{2}=0$
จะได้ $y^{2}=0$ และ $y^{4}-X^{2}y^{2}-2x^{4}=0$
$y^{2}=0$ ,$(y^{2}+x^{2})(y^{2}-2x^{2})=0$
$y^{2}=0,y^{2}=-x^{2},y^{2}=2x^{2}$ จะได้ $\frac{x^{2}}{y^{2}}=\frac{0}{0},-1,\frac{1}{2}$
จากคำถามของโจทย์คือ $(\frac{x}{y})^{4}=\frac{x^{4}}{y^{4}}=(\frac{x^{2}}{y^{2}})^{2}$
$=\frac{0}{0},1,\frac{1}{4}$ ผิดตรงไหนกรุณาบอกด้วยครับ