Where did I make a mistake solving $x^3-6x^2+2x+3=0$ with Cardano's method?

Question

Solve $x^3-6x^2+2x+3=0$ for $x$

Here's my attempt at solving it via Cardano's Method as it applies to the general cubic:

$$x^3-6x^2+2x+3=0$$ \begin{align*}y&=x+\frac{b}{3a} \\ y&= x+\frac{-6}{3} \\ y &= x-2 \\ y+2&=x\end{align*} Substitute $x$ for $y+2$ $$(y+2)^3-6(y+2)^2+2(y+2)+3=0$$ Expand out $$y^3+6y^2+12y+8-6y^2-24y-24+2y+4+3=0$$ Add like terms to get the depressed cubic $$y^3-10y-9=0$$ $$y=u+v$$ Substitute $y$ for $u+v$ $$y^3=(u+v)^3$$ \begin{align*}(u+v)^3 &= u^3+3u^2v+3uv^2+v^3 \\ &= u^3+v^3+3uv(u+v) \\ &= u^3+v^3+3uvy\end{align*} $$y^3=u^3+v^3+3uvy$$ Make the right side equal to $0$ to get the depressed cubic in terms of $u$, $v$, and $y$ $$y^3-3uvy-(u^3+v^3)=0$$ \begin{align*}-3uv &= -10 & u^3+v^3 &= 9 & uv &=\frac{10}{3}\end{align*} $$u^3v^3=\frac{10^3}{3^3}=\frac{1000}{27}$$ Factored form of desired quadratic $$(t-u^3)(t-v^3)=0$$ Expand out the quadratic \begin{align*}t^2-(u^3+v^3)t+u^3v^3 &= 0 \\ t^2-9t+\frac{1000}{27} &= 0\end{align*} Can't factor quadratic, use the quadratic formula to solve for $t$ $$t=\frac{9\pm \sqrt{81-\frac{4000}{108}}}{2}$$ Simplify the square root $$t=\frac{9 \pm \frac{\sqrt{3561}}{9}}{2}$$ Simplify the fraction $$t=\frac{81 \pm \sqrt{3561}}{18}$$ Take cube root of the entire fraction to get $u$ and $v$ \begin{align*}u &= \sqrt[3]{\frac{81 + \sqrt{3561}}{18}} \\ v &=\sqrt[3]{\frac{81 - \sqrt{3561}}{18}}\end{align*} Simplify root expression \begin{align*}u &= \frac{\sqrt[3]{972+12\sqrt{3561}}}{6} \\ v &= \frac{\sqrt[3]{972-12\sqrt{3561}}}{6} \\\\ y &= \frac{\sqrt[3]{972+12\sqrt{3561}}+\sqrt[3]{972-12\sqrt{3561}}}{6} \\\\ x &= \frac{\sqrt[3]{972+12\sqrt{3561}}+\sqrt[3]{972-12\sqrt{3561}}}{6}+2\end{align*}

Now, to check that I got this wrong, I checked for the roots of the original cubic using a calculator, and this is what I should have gotten for $x$:

$$x= \left\{1, \frac{5+\sqrt{37}}{2}, \frac{5-\sqrt{37}}{2}\right\}$$

But instead, I got a whole mess of cube roots and square roots, so I definitely made a mistake somewhere trying to solve this cubic using Cardano's Method. So where is that mistake that led to the whole mess of $n$th roots?

Answer 1

As @Macavity stated, the part where you went wrong is when you solved for the value of $t$. You multiplied both the numerator and the denominator by $4$ when in fact, only the numerator should be multiplied.

It's recommended that you first try to use the rational root theorem before resorting to this formula. You can't rely on formulas like this, especially if its degree is greater than five. However, if you really want to use Cardano's formula, it will be easier if you skip the "completing the cube" process and use a summary. Check out the formula from ProofWiki.

The solutions of the general cubic formula $$ax^3 + bx^2 + cx + d = 0$$ for $a \neq 0$ are \begin{align*}x_1 &= S + T - \frac{b}{3a} \\ x_2 &= -(S + T) - \frac{b}{3a} + \frac{i\sqrt{3}}{2}(S - T) \\ x_3 &= -(S + T) - \frac{b}{3a} - \frac{i\sqrt{3}}{2}(S - T)\end{align*} where \begin{align*}S &= \sqrt[3]{R + \sqrt{Q^3 + R^2}} \\ T &= \sqrt[3]{R - \sqrt{Q^3 + R^2}}\end{align*} and \begin{align*}Q &= \frac{3ac - b^2}{9a^2} \\ R &= \frac{9abc - 27a^2d - 2b^3}{54a^3}.\end{align*}

The expression $D = Q^3 + R^2$ is the discriminant of the equation.

If:

• $D > 0$, then one root is real and the other two are complex conjugates.
• $D = 0$, then all roots are real and at least two are equal.
• $D < 0$, then all roots are real and unequal.

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But instead, I got a whole mess of cube roots and square roots, so I definitely made a mistake somewhere trying to solve this cubic using Cardano's Method.

A messy-looking expression made up of nested roots compared to a simpler-looking expression doesn't mean that one of them is a mistake. It's just that solving expressions involving nested roots are more prone to mistakes.

Compare the following expressions: $\sqrt[3]{2} + \sqrt[3]{3}$ and $\sqrt[3]{5 + \sqrt[3]{324} + \sqrt[3]{486}}$. Although it is quite hard to see, they are equal.

In your case, your solution will eventually get to the point where $$t = \frac{9 \pm \sqrt{-\frac{1813}{27}}}{2}.$$ By letting complex numbers at play, you will have to take its cube root later. For instance, $$u = \sqrt[3]{\frac{9 \pm i\sqrt{\frac{1813}{27}}}{2}}$$ This is the so-called casus irreducibilis. Refer to this answer to know how to get the real solutions.