Quadratic Equation [2]

One approach is just to solve the quadratic (giving two complex roots), and then cubing one of the roots. This gets the answer, but it's surprisingly simple; it suggests there might be a quicker way...

Rearrange the equation to get $x^2=3x-9$ . Now, multiplying both sides of the equation by $x$ , we have $x^3=3x^2-9x$ .

Substituting $x^2=3x-9$ into this, we get $x^3=3\left(3x-9 \right) -9x =\boxed{-27}$ .

Since $\alpha$ is a root of $x^2 - 3x + 9 =0$ , then

$\begin{aligned} \alpha^2 - 3\alpha + 9 & = 0 & \small \blue{\text{Rearrange}} \\ \alpha^2 & = 3\alpha - 9 & \small \blue{\text{Multiply both sides by }\alpha} \\ \alpha^3 & = 3\blue{\alpha^2} - 9\alpha & \small \blue{\text{Note that }\alpha^2 = 3\alpha - 9} \\ & = 3\blue{(3\alpha - 9)} - 9\alpha \\ & = 9\alpha -27 - 9\alpha \\ & = \boxed{-27} \end{aligned}$

$0 = (\alpha^2 - 3\alpha + 9)(\alpha+3) = \alpha^3+27,$ so $\alpha^3 = \fbox{-27}.$