<s>[INST] How to solve $x^3 - 2x^2 -16x+16=0$? How to solve $x^3 - 2x^2 -16x+16=0$ ?
I tried to factor out $x^2$ but it doesn't work..any hints?
Ok I tried to use rational root theorem and have possible answer $\pm 1, \pm 2, \pm 4, \pm 8 , \pm 16$, but my calculator gives out some decimal numbers...
 [/INST] To solve the cubic, we first perform a transformation to eliminate the $x^2$ term:  by letting $x = y + 2/3$, we then obtain the 'depressed' cubic $$y^3 - \frac{52}{3} y + \frac{128}{27} = 0.$$  Next, recall the triple angle cosine identity:  $$4 \cos^3 \theta - 3 \cos \theta = \cos 3\theta.$$  This suggests letting $y = u \cos \theta$ so that $$\begin{align*} 0 &= u^3 \cos^3 \theta - \frac{52}{3}u \cos \theta + \frac{128}{27} \\ &= \frac{u^3}{4} \left( 4 \cos^3 \theta - \frac{208}{3u^2} \cos \theta + \frac{512}{27u^3}\right).\end{align*}$$  Thus if we chose $u = \frac{4 \sqrt{13}}{3}$, it immediately follows that $$\cos 3\theta = - \frac{8}{13\sqrt{13}}.$$  Therefore, the roots are given by $$x = \frac{2}{3} + \frac{4\sqrt{13}}{3} \cos \left( \frac{1}{3} \cos^{-1} \frac{-8}{13 \sqrt{13}} + \frac{2\pi k}{3} \right), \quad k = 0, 1, 2.$$
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