<s>[INST] How to find the root of this polynomial equation of third degree? x^3 -3x + 4 = 0 

How do I find roots of the above equation?
I have tried plugging in values of x but it is not satisfying the equation above.
 [/INST] The full solution to the cubic is:
$$x = (q + u)^{\frac{1}{3}}   +   (q - u)^{\frac{1}{3}}   +   p$$
$$u = \sqrt{q^2 + (r-p^2)^3}$$
$$p = -\frac{b}{3a}$$
$$q = p^3 + \frac{bc-3ad}{6a^2}$$
$$r = \frac{c}{3a}$$
Here we have $a=1$, $b=0$, $c=-3$, and $d=4$. Hence:
$$r = \frac{-3}{3} = -1$$
$$p = -\frac{0}{3} = 0$$
$$q = 0^3 + \frac{0-3(4)}{6} = \frac{-12}{6} = -2$$
$$u = \sqrt{(-2)^2 + (-1)^3} = \sqrt{3}$$
$$x = (\sqrt{3}-2)^{\frac{1}{3}}   +   (\sqrt{3}-2)^{\frac{1}{3}}$$
$$\therefore x= 2 (\sqrt{3}-2)^{\frac{1}{3}}$$
There are also two complex solutions, but to get these simply divide your original polynomial by the solution we get, yielding a quadratic. An application of the quadratic formula will yield the other two solutions.
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