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A popular story about the discovery of the complex numbers goes as follows. Once the formula for the solution of the cubic equation has been discovered its application to the equation $x^3=15x+4$ yields the answer $\sqrt[3]{2+\sqrt{-121}}+\sqrt[3]{2-\sqrt{-121}}$. While this is seemingly meaningless it was discovered that if one performs formal manipulations with the $\sqrt{-121}$ as if it were an ordinary number one can boil down the above expression to $4$ which is an actual soution of the above equation.

I'm interested into how one manipulates $\sqrt[3]{2+\sqrt{-121}}+\sqrt[3]{2-\sqrt{-121}}$ down to $4$ by assuming only very simple formal rules.

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Bombelli's observation or insight or wishful thinking was that $(2+i)^3 = 2 + 11i$ and so $(2-i)^3 = 2 - 11i$. Thus, $\sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i} = 2 +i + 2-i = 4$.

The stroke of genius was to assume that $\sqrt[3]{2 + 11i}=a+bi$ and that this should imply that $\sqrt[3]{2 - 11i}=a-bi$. Computing $a$ was relatively easy, as above.

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If the square root of $-1$ is called $\mathrm{i}$ we have $\sqrt{-121}=\sqrt{121}\sqrt{-1}=11 \, \mathrm{i}$. The solution is $\sqrt[3]{2+11 \, \mathrm{i}}+\sqrt[3]{2-11 \, \mathrm{i}}$ To calculate $\sqrt[3]{2+11 \, \mathrm{i}}$ we set

$$ \sqrt[3]{2+11i}=a+b \, \mathrm{i}$$

with real $a$ and $b$. By raising to the 3rd power we get $$2+11\, \mathrm{i}=(a+b \, \mathrm{i})^3$$ Now by comparing the real and imaginary part we get two equations for two variables $a$ and $b$ that can be solved to find $a=2$ and $b=1$. The same calculations can be done for $2-11\, \mathrm{i}$ to get $a=2$ and $b=-1$. I use $maxima$ to do the calculations. $\%i$ is the imaginary unit in maxima. x:expr assignes exprto the variable x. Left aligned text blocks is input to maxima, centered text is maxima output. maxima output was manually formatted by me.

/* assign to the variable z */
z:2+11*%i;

$$11 \, \mathrm{i}+2 $$

/* the cubic root of z is a complex number
with unknown real part a
and unknown complex part b*/
r:a+b*%i;

$$\mathrm{i} \, b +a $$

/* display r^3 = z */
collectterms(expand(r^3),%i)=z;

$$\mathrm{i} \, (3a^2b-b^3)-3ab^2+a^3 = 11 \, \mathrm{i} +2$$

/* from r^3=z we get an equation for the real parts
and for the imaginary parts */
eq1:realpart(r^3)=realpart(z);
eq2:imagpart(r^3)=imagpart(z);

$$a^3-3ab^2=2$$ $$3a^2b-b^3=11$$

/* we solve this system of two equations 
with the unknowns a and b*/
solve([eq1,eq2],[a,b]);

$$\begin{array}{} & [[a=2,&b=1], \\ & [a=0.93301-1.61603 \, \mathrm{i}, & b=1.06699 \, \mathrm{i}-0.61603], \\ & [a=1.61603 \, \mathrm{i} +0.93301, & b=-1.066987 \, \mathrm{i}-0.616025], \\ & [a=0.11603 \, \mathrm{i}+0.06699, & b=1.93301 \, \mathrm{i}+1.11603], \\ & [a=0.06699-0.11603 \, \mathrm{i}, & b=1.11603-1.93301 \, \mathrm{i}] ] \end{array}$$

/* now compute the other cubic root */
z:2-11*%i;

$$2 - 11 \, \mathrm{i}$$

/* display r^3 = z */
collectterms(expand(r^3),%i)=z;

$$\mathrm{i} \, (3a^2b-b^3)-3ab^2+a^3 = 2- 11 \, \mathrm{i} $$

/* from r^3=z we get an equation for the real parts
and for the imaginary parts */
eq1:realpart(r^3)=realpart(z);
eq2:imagpart(r^3)=imagpart(z);

$$a^3-3ab^2=2$$ $$3a^2b-b^3=-11$$

/* we solve this system of two equations 
with the unknowns a and b*/
solve([eq1,eq2],[a,b]);

$$\begin{array}{} &[[a=2,&b=-1], \\ &[a=1.61603 \mathrm{i} +0.93301,&b=1.06699 \mathrm{i} +0.61603],\\ &[a=0.93301-1.61603 \mathrm{i} ,&b=0.61603-1.06699 \mathrm{i} ], \\ &[a=0.06699-0.11603 \mathrm{i} ,&b=1.93301 \mathrm{i} -1.11603], \\ &[a=0.11603 \mathrm{i} +0.06699,&b=-1.93301 \mathrm{i} -1.11603]] \end{array}$$

I am only interested in real solution pairs $[a,b]$
and so the result is $(2+i)+(2-i)=4$

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    $\begingroup$ I'm not sure how to interpret your answer... more precisely how to interpret the contents of the grey box? $\endgroup$ – Timotej Oct 21 '13 at 13:49
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    $\begingroup$ I also don't understand how this is supposed to answer the question, $\endgroup$ – MJD Oct 21 '13 at 13:52
  • $\begingroup$ @Timotej: Sorry I thought the maxima languate is descriptive enough. Is it now clear how to proceed? $\endgroup$ – miracle173 Oct 21 '13 at 14:29
  • $\begingroup$ @MJD maybe it is clearer now $\endgroup$ – miracle173 Oct 21 '13 at 20:54

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