# I know there are three real roots for cubic however cubic formula is giving me non-real answer. What am I doing wrong?

I want to solve the equation $x^3-x=0$ using this cubic equation. For there to be real roots for the cubic (I know the roots are $x=-1$, $x=0$, $x=1$), I assume there must be a positive inside the inner square root. (Or is that wrong?)

However, when I substitute in $a=1$, $b=0$, $c=-1$, $d=0$, the square root term inside the cube root terms becomes

$$\sqrt{\;\left(\;2(0)^3 - 9(1)(0)(-1) + 27(1)^2(0)\;\right)^2 - 4 \left(\;(0)^2 - 3(1)(-1)\;\right)^3\quad}$$

It gives me $\sqrt{-108}$, which is $10.39i$. Now that I have a non-real number as part of the equation I can't see any way for it to be cancelled or got rid of, even though I know there is a real answer.

Could somebody please tell me how I can get a real answer and what I am doing wrong? Thanks.

• Is there a particular reason as to why you want to use that formula? – fosho Jan 6 '16 at 19:10
• I'm trying to find a general formula which I'm using in a more complex question, I just used x^3-x=0 as an example where I couldn't get it to work – Alex Jones Jan 6 '16 at 19:20
• Please look at the Wikipedia article on the Casus Irreducibilis. – André Nicolas Jan 6 '16 at 19:31

Okay, so you're getting that operands of the cube root parts of the formula will look like this: $$p + q i \qquad\text{and}\qquad p - q i$$ with some pesky non-zero $q$ (namely, $\sqrt{108}$). Well, these values are conjugates, so that their respective (principal) cube roots are conjugates, as well. For the $x_1$ value in your formula, these conjugate cube roots add together, and their imaginary parts conveniently cancel. The same kind of cancellation happens for the $x_2$ and $x_3$ values, too, because the factors $\frac{1}{2}(1+i\sqrt{3})$ and $\frac{1}{2}(1-i\sqrt{3})$ are themselves conjugates. When the dust settles, you'll have the three real roots you expect.