x + y + z = 0 [given] On squaring both sides, we get (x+y+z)2=0 ⇒x2+y2+z2+2(xy+yz+zx)=0 ⇒x2+y2+z2=−2(xy+yz+zx) Again, squaring both sides, we get (x2+y2+z2)2=4(xy+yz+zx)2 ⇒x4+y4+z4+2(x2y2+y2z2+x2z2) = 4[x2y2+y2z2+z2x2+2xyz(x+y+z)] ⇒x4+y4+z4=2(x2y2+y2z2+x2z2) ∴