x²+4x+3=(x+1)(x+3)
x³+27 =(x+3)(x²-3x+9)
x²-3x+2=(x-1)(x-3)
∵(x-1) / (x²+4x+3) 与 (x³+27) / (x²-3x+2) 互为倒数
∴[(x-1) / (x²+4x+3)] × [(x³+27) / (x²-3x+2)] = 1
∴(x-1)(x³+27) = (x²+4x+3)(x²-3x+2)
∴(x-1)(x+3)(x²-3x+9) = (x+1)(x+3)(x-1)(x-3)
∴x²-3x+9 = (x+1)(x-3)
∴x²-3x+9 = x²-2x-3
∴x=12