【答案】应助回帖
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fengfeng子: 金币+10, 博学EPI+1, ★★★很有帮助, 你这个是自己导出来的还是本身就有这个万能公式？ 2015-10-09 21:04:39
与上面的图片对照的看
{{x = 1/3 - (2^(1/3) (-1 + 3 a - 3 b - 3 b^2))/(3 (2 - 9 a + 27 ab + 9 b + 9 b^2 +
Sqrt[4 (-1 + 3 a - 3 b - 3 b^2)^3 + (2 - 9 a + 27 ab + 9 b +9 b^2)^2])^(1/3)) + (1/( 3 *2^(1/3)))
((2 - 9 a + 27 ab + 9 b + 9 b^2 + Sqrt[4* (-1 + 3 a - 3 b - 3 b^2)^3 + (2 - 9 a + 27 ab + 9 b + 9b^2)^2])^(1/3))}；
{x = 1/3 + ((1 + I Sqrt[3]) (-1 + 3 a - 3 b - 3 b^2))/(3* 2^(2/3) (2 - 9 a + 27 ab + 9 b + 9 b^2 +
Sqrt[4* (-1 + 3 a - 3 b - 3 b^2)^3 + (2 - 9 a + 27 ab + 9 b +9 b^2)^2])^(1/3)) - (1/(6* 2^(1/3)))(1 - I Sqrt[3]) (2 - 9 a + 27 ab + 9 b + 9 b^2 + Sqrt[4*(-1 + 3 a - 3 b - 3 b^2)^3 + (2 - 9 a + 27 ab + 9 b +
9b^2)^2])^(1/3)}；
{x =1/3 + ((1 - I Sqrt[3]) (-1 + 3 a - 3 b - 3 b^2))/(3 2^(2/3) (2 - 9 a + 27 ab + 9 b + 9 b^2 +
Sqrt[4* (-1 + 3 a - 3 b - 3 b^2)^3 + (2 - 9 a + 27 ab + 9 b + 9 b^2)^2])^(1/3)) - (1/(6*2^(1/3)))(1 + I Sqrt[3]) (2 - 9 a + 27 ab + 9 b + 9 b^2 + Sqrt[4*(-1 + 3 a - 3 b - 3 b^2)^3 + (2 - 9 a + 27 ab + 9 b +
9b^2)^2])^(1/3)}} |
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