Answer :
Cubic equation / cubic polynomial : x³ - 3 p x - 2 q = 0
Solution is given by:
[tex]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}] [/tex]
Or, by:
[tex]x = [q+\sqrt{q^2-p^2}]^{\frac{1}{3}}+[q-\sqrt{q^2-p^3}]^{\frac{1}{3}}[/tex]
Given y³ - 7 y + 6 = 0
So substituting p = 7/3 and q = -3 we get :
[tex]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[/tex]
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 :
[tex]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}}[/tex]
This solution gives complex numbers. So it requires to convert them in the polar coordinate form [tex]e^{i\theta}[/tex] and then take cube root and then add the two cube roots.
Then we get the same answers.
Solution is given by:
[tex]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}] [/tex]
Or, by:
[tex]x = [q+\sqrt{q^2-p^2}]^{\frac{1}{3}}+[q-\sqrt{q^2-p^3}]^{\frac{1}{3}}[/tex]
Given y³ - 7 y + 6 = 0
So substituting p = 7/3 and q = -3 we get :
[tex]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[/tex]
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 :
[tex]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}}[/tex]
This solution gives complex numbers. So it requires to convert them in the polar coordinate form [tex]e^{i\theta}[/tex] and then take cube root and then add the two cube roots.
Then we get the same answers.