As part of a program I'm writing, I need to solve a cubic equation exactly (rather than using a numerical root finder):

```
a*x**3 + b*x**2 + c*x + d = 0.
```

I'm trying to use the equations from here. However, consider the following code (this is Python but it's pretty generic code):

```
a = 1.0
b = 0.0
c = 0.2 - 1.0
d = -0.7 * 0.2
q = (3*a*c - b**2) / (9 * a**2)
r = (9*a*b*c - 27*a**2*d - 2*b**3) / (54*a**3)
print "q = ",q
print "r = ",r
delta = q**3 + r**2
print "delta = ",delta
# here delta is less than zero so we use the second set of equations from the article:
rho = (-q**3)**0.5
# For x1 the imaginary part is unimportant since it cancels out
s_real = rho**(1./3.)
t_real = rho**(1./3.)
print "s [real] = ",s_real
print "t [real] = ",t_real
x1 = s_real + t_real - b / (3. * a)
print "x1 = ", x1
print "should be zero: ",a*x1**3+b*x1**2+c*x1+d
```

But the output is:

```
q = -0.266666666667
r = 0.07
delta = -0.014062962963
s [real] = 0.516397779494
t [real] = 0.516397779494
x1 = 1.03279555899
should be zero: 0.135412149064
```

so the output is not zero, and so x1 isn't actually a solution. Is there a mistake in the Wikipedia article?

ps: I know that numpy.roots will solve this kind of equation but I need to do this for millions of equations and so I need to implement this to work on arrays of coefficients.

Wikipedia's notation `(rho^(1/3), theta/3)`

does not mean that `rho^(1/3)`

is the real part and `theta/3`

is the imaginary part. Rather, this is in polar coordinates. Thus, if you want the real part, you would take `rho^(1/3) * cos(theta/3)`

.

I made these changes to your code and it worked for me:

```
theta = arccos(r/rho)
s_real = rho**(1./3.) * cos( theta/3)
t_real = rho**(1./3.) * cos(-theta/3)
```

(Of course, `s_real = t_real`

here because `cos`

is even.)

I've looked at the Wikipedia article and your program.

I also solved the equation using Wolfram Alpha and the results there don't match what you get.

I'd just go through your program at each step, use a lot of print statements, and get each intermediate result. Then go through with a calculator and do it yourself.

I can't find what's happening, but where your hand calculations and the program diverge is a good place to look.

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