∵ x²+2x-2=0
∴ x²+2x=2
原式=[(x²-4)/(x²+x+1)]²÷[(x³-2x²)/(x³+x²+x)]²×x/(x+2)³
=[(x+2)(x-2)/(x²+x+1)]²÷{x²(x-2)/[x(x²+x+1)]}²×x/(x+2)³
=[(x+2)²(x-2)²/(x²+x+1)²]÷{x^4(x-2)²/[x²(x²+x+1)²]}×x/(x+2)³ ( ^ 表示乘方 )
=[(x+2)²(x-2)²/(x²+x+1)²]×{x²(x²+x+1)²/[x^4(x-2)²]}×x/(x+2)³
=1/[x(x+2)]
=1/(x²+2x)
=1/2