Bombelli's wild thought of cubic equations

Question

In many books, like Visual Complex Analysis. talk about the real original of complex number. the author begin with this equation:

$$x^3=15x+4$$

Rewrite to:

$$x^3-15x-4=0$$

So that $p= -15$ and $q=-4$.

Then the author use the formula $$x=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$$

to say that the equation has a root $$x = \sqrt[3]{2+11i}+\sqrt[3]{2-11i}$$

Apparently, $x=4$ is a root of the equation $x^3=15x+4$. Then the author guess $$\sqrt[3]{2+11i}+\sqrt[3]{2-11i} =4.$$ then introduction complex arithmetic. It seems very natural. but equation $x^3=15x+4$ has three different roots. why the author, or probably Bombelli guess this equal $4$, not other roots?

Answer 1

He (Bombelli) probably used the Cardano formula to obtain the $\textbf{one}$ root solution you just described $$x = \sqrt[3]{2+11i}+\sqrt[3]{2-11i}$$

He then had to find the cube roots of each of the radicals above. I am not sure exactly how he found them but he may have used the following sometimes used polynomial to find each of the cube roots
$$\frac{-64a^9+(48x)a^6+((15(x)^2)-3(3y)^2)a^3+(x)^3}{-64} = 0$$ where $x$ = $2$ and $y$ = $11i$ to get

$$a^9 + \frac{3}{2}a^6 - \frac{3327}{64}a^3 + \frac{1}{8} = 0$$

This polynomial has one rational root $a$ = $2$

He could now take the rational root $a$ = $2$ he found above to find $b$ in the equation below in order to denest the two cube roots in Cardano's formula. The $2$ on the $RHS$ is the $2$ under the radical $$ a^3+3ab^2=2$$

$$2^3 + 3(2)b^2 = 2$$ $$b=\pm \sqrt\frac{-6}{6} = \pm i$$

So using the $a$ he found and the $b$ he found he could derive the cube roots as $a + b$. This gives the two results $$2 + i$$ and $$2 -i$$ Each is the cube root of the nested radicals in the Cardano solution.

So using the $a$ he found and the $b$ he found he could observe that this was equal to the real root of $4$ found using other methods

$$\left(2 + i\right) + \left(2 - i\right) = 4$$

Answer 2

$(2\pm i)^3=8\pm 3*4i-3*2\pm i=2\pm 11i$. Therefore $\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=2+i+2-i=4.$

Answer 3

First of all, the other two roots may be found by multiplying the cube roots of the solution with primitive cube of unity, which was not discoverd yet. The other two real solutions $x = -\sqrt{3}-2$ and $x=\sqrt{3}-2$ are both negative. From a geometric perspective this would be useless (Cardano even said so himself).

$\mathbf{Fun fact}$: In our calculations we use $i$, but that notation was not to be introduced for another 200 years by none other than Leonard Euler.

The thing Bombelli did was converting $\sqrt[3]{2+11i}$ into $2+i$, so the imaginary parts of the solution negate eachother. He imagined that part of the solution $\sqrt[3]{2+\sqrt{-121}}$ obtained from Cardano's method can be rewritten as something like $a+ b\sqrt{-1}$, which is the genius part. Ofcourse he did not know in advance if this would work, however since there is an obvious solution $x=4$ there was certainly hope that it would.

It seems unlikely he would've started the general expression $\sqrt[3]{x+yi}$ here. We will do so however:

$$\sqrt[3]{x+yi} = a+bi$$

Raise $LHS$ and $RHS$ to power of 3:

$$\begin{align} x+yi & = (a+bi)^3 \\ x+yi & = a^3-3ab^2 + (3a^2b-b^3)i \end{align}$$

This yields two equations:

$$a^3-3ab^2 = x \qquad (1)$$

$$3a^2b-b^3 = y \qquad (2)$$

From $(1)$ we obtain:

$$ b = \sqrt{\large\frac{a^3-x}{3a}} \quad \lor\quad b = -\sqrt{\large\frac{a^3-x}{3a}} \qquad (3)$$

Substituting the left expression in $(2)$ gives us:

$$ \large(3a^2- \large\frac{a^3-x}{3a})\cdot\sqrt{\large\frac{a^3-x}{3a}} = y$$

Merging the two fractions in the brackets:

$$\large(\large\frac{8a^3-x}{3a})\cdot\sqrt{\large\frac{a^3-x}{3a}} = y$$

Raise both sides of the equation to the power of 2 to get:

$$ \large\frac{x^2 +16a^3x+64a^6}{9a^3}\cdot\large\frac{a^3-x}{3a} = y^2$$

This can be simplified to the polyonmial:

$$ 64a^9 -48a^6x-(15x^2+27y^2)a^3-x^3= 0$$

Now we fill in $x=2$ and $y=11$ from $\sqrt[3]{2+11i}$:

$$ 64a^9 -96a^6-3327a^3-8= 0$$

Substitute $a^3$ for $x$ gives as the cubic equation:

$$ 64x^3 -96x^2-3327x-8= 0 \qquad (4)$$

Using Cardano's method we can turn this into a depressed cubic of the form $t^3+pt+q$, where

$$ p=\large\frac{3ac-b^2}{3a^2} = -\frac{3375}{64}$$

$$ q=\large\frac{2b^3-9abc+27a^2d}{27a^3}=-\frac{3375}{128}$$

In the two formula's above $a$, $b$, $c$ and $d$ represent the coeficients of $(4)$, not to be confused with the variables $a$ and $b$ from $(1)$. Solving $t^3+px+q = 0$ using Cardano's formula yields:

$$ t=7.5$$

I used Wolfram Alpha for the calculation above.

Calculate $x$:

$$ x = t-\frac{b}{3a} = 7.5 - \large\frac{-96}{3\cdot64} = 8$$

To understand this bit above check out how to create a depressed cubic from a cubic equation. I recommend the reader to watch the informative YouTube video of Derek Muller's Veritasium channel: https://www.youtube.com/watch?v=cUzklzVXJwo&t=1022s Start at time 10:55.

Next calculate $a$:

$$ a =\sqrt[3]{x} = \sqrt[3]{8} = 2$$

Finally find $b$ using the left of our expressions of $(3)$ for $a=2$ and $x=2$:

$$ b = \sqrt{\large\frac{a^3-x}{3a}} = \sqrt{\large\frac{2^3-2}{3\cdot 2}} = 1$$

Note that we can not use the right, since this will give you $b=-1$ for $y=11$, which is incorrect.

Now we know $a=2$ and $b=1$ in case of $x=2$ and $y=11$ and we may conclude that:

$$ \sqrt[3]{2+11i} = 2+i$$

A similar prove for $y=-11$ leads to:

$$ \sqrt[3]{2-11i} = 2-i$$

At some point using Cardano's method Bombelli would have had to calculate:

$$ \begin{align} \Delta & = \large\frac{q^2}{4}+\frac{p^3}{27}=\frac{\large(-\frac{3375}{128})^2}{4}+\frac{\large(-\frac{3375}{64})^3}{27} \\ & = \large\frac{\large(\frac{11390625}{16384})}{4}+\frac{\large(-\frac{38443359375}{262144})}{27} = \large-\large\frac{1378265625}{262144} \end{align} $$

Apart from the fact that it seemes unlikely for Bombelli to have performed these calculations the more convincing argument is that $\Delta < 0$. He would have had tot calculate the squareroot of a negative number again, which was kinda the thing he was trying to achieve in the first place. This proves he did not use Cardano's method, but rather guessed the values of $a$ and $b$.

It would be more likely he found $a$ and $b$ through equations $(1)$ and $(2)$ by guessing them or just $a$ through equation $(4)$.

Also note that if you express $a$ in terms of $b$ in step $(3)$ you obtain the equation:

$$ 64b^9+528b^6-1923b^3+1331 = 0$$

Now all you need to do is notice that the sum of the coefficients equal $0$. Thus $b=1$ is a solution. This is perhaps easier than obtaining the number by guessing them from $(1)$ an $(2)$ (with $x=2$ and $y=11$ ofcourse). Thus I deem this the most likely scenario to have happened.