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Answer to Question #4515 in Algebra for Inga

Question #4515
How to get exact solutions of equation x^3+3x^2-1=0.
The solution can't be decimal approximation, it should be analytical solution.
Expert's answer
x^3+3x^2-1=0
We use here trigonometric Viete formula
for equation written in form x^3+a x^2+b x+c=0.

Q=(a^2-3b)/9
R=(2a^3-9ab+27c)/54
S=Q^3-R^2

For our equation Q=(3^2-3*0)/9=1
R=27/54=1/2
S=3/4

fi=1/3*arccos(R/Q3/2)
x1=-2sqrt(Q) cos fi-a/3
x2=-2sqrt(Q) cos( fi+2 pi/3)-a/3
x3=-2sqrt(Q) cos( fi-2 pi/3)-a/3

for our data
fi=1/3 arccos(0.5)
x1=-2 cos(1/3*arccos(0.5))-1
x2=-2 cos(1/3*arccos(0.5)+2 pi/3)-1
x3=-2 cos(1/3*arccos(0.5)-2 pi/3)-1
where x1, x2, x3 - the roots of our equation

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