Here's my stab.
I took your equation and expanded it to;
(x^6 + 30x - 1) / x^3 = 0 ( I don't feel like typing out the steps, I'm sure you can do it too.)
So this only equals zero if;
x^6 + 30x - 1 = 0
Let x^3 = y, then;
y^2 + 30y - 1 = 0
Using the quadratic equation gives;
y = .033296378.. and -30.03329638...
Since y = x^3, x = y^1/3
x = .321710818... and -3.1081632...
I checked this with my TI-89 and my answer was confirmed correct. The slight error from being exact is only from rounding the value of x.
The precise answer is;
-(√(226) + 15)^(1/3) and (√(226) - 15)^(1/3)
Which agrees with what I had earlier.
We don' need no stinkin' calculators....
Last edited by irspow (2006-02-07 12:05:40)