Exception on the Cubic Formula

Question

I have searched for the cubic formula, which is: $$ \sqrt[3]{\frac{-B^3}{27A^3} + \frac{BC}{6A^2} - \frac{D}{2A} + \sqrt{\left(\frac{-B^3}{27A^3} + \frac{BC}{6A^2} - \frac{D}{2A}\right) ^ 2 + \left(\frac{C}{3A} - \frac{B^2}{9A^2}\right)^3}} + \sqrt[3]{\frac{-B^3}{27A^3} + \frac{BC}{6A^2} - \frac{D}{2A} + \sqrt{\left(\frac{-B^3}{27A^3} + \frac{BC}{6A^2} - \frac{D}{2A}\right) ^ 2 -\left(\frac{C}{3A} - \frac{B^2}{9A^2}\right)^3}} - \frac{B}{3A}$$ However, when I substitute A = 1, B = 6, C = 11, D = 6, then I should expect something like this: $$ x^3 + 6x^2 + 11x + 6 = 0 $$ $$ x = -1 \, or \, x = -2 \, or \, x = -3 $$ However I have got complex solution. What is happening?

Answer 1

You made a mistake while copying your formula. In your second cube root, there should be a minus sign before the square root and a plus sign inside the square root where you now have a minus sign. Therefore, you'll see that the cube roots cancel and you'll be left with the last term which is -2.

The correct formula:

$$\sqrt[3]{\frac{-B^3}{27A^3} + \frac{BC}{6A^2} - \frac{D}{2A} + \sqrt{(\frac{-B^3}{27A^3} + \frac{BC}{6A^2} - \frac{D}{2A}) ^ 2 + (\frac{C}{3A} - \frac{B^2}{9A^2})^3}} + \sqrt[3]{\frac{-B^3}{27A^3} + \frac{BC}{6A^2} - \frac{D}{2A} - \sqrt{(\frac{-B^3}{27A^3} + \frac{BC}{6A^2} - \frac{D}{2A}) ^ 2 + (\frac{C}{3A} - \frac{B^2}{9A^2})^3}} - \frac{B}{3A}$$