given that x=(2x^2-x+8)y
show that (x-1+y)^2/4y=(1+2y-63y^2)/16y^2
odd definition, but try just substituting in for y:
y = x/(2x^2-x+8)
(x-1+y)^2/4y
=(x-1+(x/(2x^2-x+8)))^2/(4(x/(2x^2-x+8)))
= (2x^3-3x^2+10x-8)^2 / 4x(2x^2-x+8)
(x-1+y)^2/4y=(1+2y-63y^2)/16y^2
=(1+2(x/(2x^2-x+8))-63(x/(2x^2-x+8))^2)/(16(x/(2x^2-x+8))^2)
= (x^2-4)^2 / 4x^2
I don't think they are equal. Plug in 1 for x
1 = 9y, so y = 1/9
1/36 ≠ (1 + 2/9 - 63/81)/(16/81) = 9/4