解:两边立方, 得
(x³-x-3)³=8(6x-3x²)
x⁹-3x⁷-9x⁶+3x⁵+18x⁴+26x³+15x²-75x-27=0
因式分解, 得
(x³-3x-1)(x⁶-8x³+3x²-6x+27)=0
(x³-3x-1)[(x³-4)²+3(x-1)²+8]=0
∴x³-3x-1=0, 令2t=x, 得
8t³-6t=1
4t³-3t=1/2 由余弦的三倍角公式
==> cos3θ=1/2 ==> 3θ=π/3, 5π/3, 7π/3
==> θ=π/9, 5π/9, 7π/9
==> t=cos(π/9), cos(5π/9), cos(7π/9)
即x=2cos(π/9), 2cos(5π/9), 2cos(7π/9)。
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