Question

y^3-7y+6=?
How solve this

Answer 1

Cubic equation / cubic polynomial  :  x³ - 3 p x - 2 q = 0
Solution is given by:     
$x_k,\ \ k=0,1,2\\\\x_k=2\ *\ \sqrt{p}\ *\ Cos[\frac{1}{3}\ *\ Cos^{-1}(\frac{q}{p^{\frac{3}{2}}})-\frac{2\pi k}{3}]$

Or, by:
$x = [q+\sqrt{q^2-p^2}]^{\frac{1}{3}}+[q-\sqrt{q^2-p^3}]^{\frac{1}{3}}$

Given  y³ - 7 y + 6 = 0
So substituting      p = 7/3    and    q = -3  we get :

$x_0=2*\sqrt{7/3}\ Cos[\frac{1}{3}*Cos^{-1}(\frac{-3}{(7/3)^{\frac{3}{2}}})-\frac{2\pi *0}{3}]=2\\\\x_1=2*\sqrt{7/3}\ Cos[\frac{1}{3}*Cos^{-1}(\frac{-3}{(7/3)^{\frac{3}{2}}})-\frac{2\pi *1}{3}]=1\\\\x_2=2*\sqrt{7/3}\ Cos[\frac{1}{3}*Cos^{-1}(\frac{-3}{(7/3)^{\frac{3}{2}}})-\frac{2\pi *1}{3}]=-3$

We can try to find the solutions of  the polynomial equation: y³ - 7 y + 6 = 0  by
  constant term / coefficient of y³:    +or -  6 /1  or its factors :  + or - of 6, 3, 2, 1
So we can try these 8 possibilities.  It so happens that 1, 2, -3 are the roots.

By using the other formula :

$x = [ -3 + \sqrt{(-3)^2-(7/3)^3} ]^{\frac{1}{3}} + [q - \sqrt{(-3)^2-(7/3)^{\frac{1}{3}}}]\\\\=[-3+1.9245\ i]^{\frac{1}{3}}+[-3-1.9245\ i]^{\frac{1}{3}}$

This solution gives complex numbers.  So it requires to convert them in the polar coordinate form  $e^{i\theta}$  and then take cube root and then add the two cube roots.
Then we get the same answers.