True or False : Every root of a cubic equation has closed expression

Question

I know that statement is right in many examples but when I wanted to find exact value of $10$ degree , I achieved this equation : $8t^3 - 6t + 1 = 0$ where $t = \sin10$ . Also I know we have a nested radical expression for it (See :http://mathworld.wolfram.com/TrigonometryAnglesPi18.html ) And contradiction appears when it's a root of equation and we can also solve it by Cardano's method and this gives us a closed formula for $\sin 10$.

Answer 1

Taking $$ x = 2 \sin \left( \frac{\pi}{18} \right) \approx 0.347296 > \frac{1}{3} $$ we are told $$ x = \sqrt{2 - \sqrt{2 + \sqrt{2 + x} }}. $$ Put another way, we are demanding a fixpoint of the real valued function $$ f(w) = \sqrt{2 - \sqrt{2 + \sqrt{2 + w} }}. $$

I repeatedly squared and put the polynomial terms on one side, which led to $$ x^8 - 8 x^6 + 20 x^4 - 16 x^2 - x + 2 = 0. $$ This factors, $$ \left(x-2 \right) \left( x+1 \right) \left(x^3 + x^2 - 2x+1 \right) \left( x^3 - 3x + 1\right) $$ You can see from the graphs that $x \approx 0.347$ can only be a root of the final cubic, $x^3 - 3x + 1.$ Call your quantity $s$ so that $x = 2s,$ we get $8 s^3 - 6 s + 1 = 0.$

Notice that, while the octic polynomial has eight real roots, the original function has just a single real fixpoint.

Answer 2

Here's a good one. Find the roots of $$ x^3 + x^2 - 4 x + 1. $$ Very trigonometric.