<s>[INST] Solving $x^5+x^4-12x^3-21x^2+x+5=0$ $$x^5+x^4-12x^3-21x^2+x+5=0$$ 
I think it can be solved by trigonometric ways, but how?
 [/INST] This is an old question, but should be interesting to answer. If you want an aesthetic version, then the solution to,
$$x^5 + x^4 - 12x^3 - 21x^2 + x + 5 = 0$$
is given by,
$$x = \frac{1}{5}\left(-1+z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}\right)$$
and the $z_i$ are the roots of the quartic,
$$z^4 + 12679 z^3 + 78678031 z^2 + 362989005529 z + 31^{10} = 0$$
Notice how the quartic has a constant term that is a fifth power. This factors over the extension $\sqrt{5}$. So alternatively,
$$x = \frac{1}{5}\left(\beta_0-1+\frac{31}{\beta_0}+\frac{31}{\alpha}+\alpha\right)$$
where,
$$\alpha = \left(\tfrac{31}{4}\right)^{1/5}\left(-409-125\sqrt{5}+5\sqrt{-10(925-409\sqrt{5})}\right)^{1/5}\tag1$$
$$\beta_0 = \left(\tfrac{31}{4}\right)^{1/5}\left(-409+125\sqrt{5}+5\sqrt{-10(925+409\sqrt{5})}\right)^{1/5}\tag2$$

Added:

Israel's Maple answer can be simplified. The five roots $x_k$ for $k=0,1,2,3,4$ are,
$$x_k = \frac{1}{5}\left(\frac{1}{\beta_k^{-1}}-\frac{1}{\beta_k^0}+\frac{31}{\beta_k^1}+\frac{31a}{\beta_k^2}+\frac{31b}{\beta_k^3} \right)$$
where,
$$a=\frac{\beta_0^2}{\alpha}=\tfrac{1}{4}\left(11+5\sqrt{5}+\sqrt{-10(25-11\sqrt{5})}\right)$$
$$b=\frac{\alpha\,\beta_0^3}{31}=\tfrac{1}{4}\left(-1-5\sqrt{5}+\sqrt{-10(1525-\sqrt{5})}\right)$$
$$\beta_k = e^{2\pi\,i\,k/5}\,\left(31ab\right)^{1/5}\tag3$$
So it turns out that the root $(2)$ can be factored as $(3)$. Maple's answer also has the advantage that only one fifth root extraction of a complex number is needed.
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