# Cardano's Formulas help

I am working on solving this cubic: $x^3 +x^2 - 2 = 0$ using Cardano's explicit formulas: $$A = \sqrt[3]{{-27\over 2}q + {3 \over 2} \sqrt{-3D}} \qquad B = \sqrt[3]{{-27\over 2}q - {3 \over 2} \sqrt{-3D}}$$ where for a negative discriminant, I will get a real root $\alpha = {A+ B \over 3}$ for $p,q,D$ defined as follows:

Start with an equation of the form $x^3 + ax^2 + bx + c = 0$, transform using $y = x-a/3$ to get an equation of the form $y^3 + py + q = 0$, and the discriminant of the cubic: $D = -4p^3 - 3q^2$.

My Work

I transformed the above cubic to $y^3 - {1\over 3}y - {52\over 27}$ using the substitution $y = x-1/3$. I then got a discriminant of $D = -{100}$. However, when I plug this value of $D$ into my formulas for $A$ and $B$, I get $A = \sqrt[3]{26 + 15\sqrt 3}, B = \sqrt[3]{26 - 15\sqrt 3}$. And I know that one of the roots is given by $\alpha = {A+ B \over 3}$

$(2 + \sqrt 3)^3 = 26 + 15\sqrt 3$

So my formula for $\alpha = {A+ B \over 3} = {4\over 3}$. But I am supposed to check that this equation has the real root

${1\over 3} (\sqrt[3]{26 + 15\sqrt 3} + \sqrt[3]{26 - 15\sqrt 3} \mathbf{ - 1})$

That would make sense because then it would give me a real root of $1$ which can be checked directly to be a root of this cubic. I suppose my question is: where does the $1$ come from?

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It should be $y=x+\frac{1}{3}$. Then after we have found a solution $y_0$ of the cubic in $y$, the corresponding $x_0$ is $y_0-\frac{1}{3}$. That's where the $-1$ comes from.