$x^3-6x^2+2x-2=0$ solve for x

Question

I solved it by approximation but its very long way to solve, so i was wondering if there is a quicker way to solve this kind of questions.

Answer 1

You could just use General cubic formula

Or for another way:

$$x^3-6x^2+2x-2=0$$ $$\implies (x-2)^3 - 10(x-2) -14 = 0$$

Let $t = x - 2$, then:

$$t^3 - 10t -14 = 0$$

And then apply Cardano's formula

Either way, the result will just be:

$$x = 2+\sqrt[3]{\dfrac{63+\sqrt{969}}{9}}+\sqrt[3]{\dfrac{63-\sqrt{969}}{9}}$$

Answer 2

As an addition to @The 2nd's excellent answer, you can ask a computer to find the solution for you. It is very fast. Here is an exemple in the Python language with the sympy module. It gives you a list of the three complex solutions

>>> from sympy.solvers import solve
>>> from sympy import Symbol
>>> x = Symbol('x')
>>> solve(x**3-6*x**2+2*x-2, x)
[2 + (-1/2 - sqrt(3)*I/2)*(sqrt(969)/9 + 7)**(1/3) + 10/(3*(-1/2 - 
sqrt(3)*I/2)*(sqrt(969)/9 + 7)**(1/3)), 2 + 10/(3*(-1/2 + sqrt(3)*I/2)* 
(sqrt(969)/9 + 7)**(1/3)) + (-1/2 + sqrt(3)*I/2)*(sqrt(969)/9 + 7)**(1/3), 
10/(3*(sqrt(969)/9 + 7)**(1/3)) + 2 + (sqrt(969)/9 + 7)**(1/3)]

Answer 3

Don't you prefer the nice $$x=2 \left(1+\sqrt{\frac{10}{3}} \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{21 }{10}\sqrt{\frac{3}{10}}\right)\right)\right)$$ that you could easily obtain using the steps given in the Wikipeadia page.