A complete cubic

When the cubic is not depressed, one first applies the substitution that removes the second-degree term and then Cardano’s formula.

Example — Complete cubic: $x^3-3x^2-3x-5=0$

I set $x=y+1$ and depress. Expanding: $(y+1)^3-3(y+1)^2-3(y+1)-5 = y^3-6y-10 = 0.$ $p=-6$, $q=-10$, $D=25-8=17>0$. Formula: $y=\sqrt[3]{5+\sqrt{17}}+\sqrt[3]{5-\sqrt{17}}\approx 2{,}904.$ Hence $x = y+1\approx 3{,}904$ is the real root.

Links

Topics: Second-degree equations
Concepts: Third-degree equation · Depressed form
Methods: Cardano cubic
Skills: Solving equations