[กิตติ]

ขอลองแก้ดู
$x^4+y^4=56$......(1)
$x-y=2$..........(2)
ให้$x+y=a$........(3)
$x=\frac{a+2}{2} ,y=\frac{a-2}{2} $
(2)x(3) $x^2-y^2=2a$
$(x^2-y^2)^2 = 4a^2$
$x^4+y^4-2(xy)^2 =4a^2$
$56-2(xy)^2 =4a^2$
$xy = \frac{a^2-4}{4} $...แทนลงไป
$56-2(\frac{a^2-4}{4})^2 =4a^2$
$28-(\frac{a^4-8a^2+16}{16}) = 2a^2$
$448-a^4+8a^2-16 = 32a^2$
$a^4-24a^2-432=0$
$a^2=36,-12 $
$a=\pm 6 ,a=\pm 2\sqrt{3}i $

$x=4,y=2$
$x= -2,y= -4$
$x=1+\sqrt{3}i,y= -1+\sqrt{3}i$
$x=1-\sqrt{3}i,y= -1-\sqrt{3}i$