Why is the sum of the roots of $x^3+9=12x$ zero?

Question

I found the roots of the equation to be

$$x=3,\frac{\sqrt{21}}{2}-\frac{3}{2},-\frac{\sqrt{21}}{2}-\frac{3}{2}$$

Algebraically, why is the sum of these roots equal to zero?

Answer 1

Hint: the general form $$x^3-(x_1+x_2+x_3)x^2+(x_1x_2+x_2x_3+x_3x_1)x-x_1x_2x_3=0$$

Answer 2

Consider the following cubic form:

$$ax^2+bx^2+cx+d=0$$

For your equation:

$$1\cdot x^3+0\cdot x^2-12\cdot x+9=0$$

Using Vieta's formulas, the sum of roots is equal to $-\frac{b}{a}$.

$$-\frac{0}{1}=0$$

Thus the sum of roots is $0$.

Answer 3

Consider a monic polynomial $f(x)$ of degree three with three roots: $a_0$, $a_1$, $a_2$. So $$f(x) = (x - a_0)(x - a_1)(x - a_2).$$ Expanding gives $$f(x) = x^3 - (a_0 + a_1 + a_2) x^2 + (\dots) x + (\dots).$$ So, the coefficient of $x^2$ is the negation of the sum of the roots.

In your case, the coefficient of $x^2$ is $0$, so the sum of the roots is $0$.