Author 
Comment/Response 
yehuda benshimol

09/22/09 08:31am
I'm sure you already resolved this issue
but in any case use the function N
{{x > (1/3) + 25/(3 (1/2 (101 + 3 I Sqrt[5811]))^(1/3)) +
1/3 (1/2 (101 + 3 I Sqrt[5811]))^(1/3)}, {x > (1/3) 
1/6 (1 + I Sqrt[3]) (1/2 (101 + 3 I Sqrt[5811]))^(1/3)  (
25 (1  I Sqrt[3]))/(
3 2^(2/3) (101 + 3 I Sqrt[5811])^(1/3))}, {x > (1/3) 
1/6 (1  I Sqrt[3]) (1/2 (101 + 3 I Sqrt[5811]))^(1/3)  (
25 (1 + I Sqrt[3]))/(3 2^(2/3) (101 + 3 I Sqrt[5811])^(1/3))}} //N
or
N[{{x > (1/3) + 25/(3 (1/2 (101 + 3 I Sqrt[5811]))^(1/3)) +
1/3 (1/2 (101 + 3 I Sqrt[5811]))^(1/3)}, {x > (1/3) 
1/6 (1 + I Sqrt[3]) (1/2 (101 + 3 I Sqrt[5811]))^(1/3)  (
25 (1  I Sqrt[3]))/(
3 2^(2/3) (101 + 3 I Sqrt[5811])^(1/3))}, {x > (1/3) 
1/6 (1  I Sqrt[3]) (1/2 (101 + 3 I Sqrt[5811]))^(1/3)  (
25 (1 + I Sqrt[3]))/(3 2^(2/3) (101 + 3 I Sqrt[5811])^(1/3))}} ]
yehuda
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