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When solving the following cubic equation:

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

I got one of the solutions:

$$x = \sqrt[3]{2 {\color{red}+} 11i} + \sqrt[3]{2 {\color{red}-} 11i}$$

When I calculated it with a hand calculator, it turned out to be exactly $4$. And indeed, when I substitute $x=4$ into the original equation, it is a solution. So this appears to be true:

$$\sqrt[3]{2 {\color{red}+} 11i} + \sqrt[3]{2 {\color{red}-} 11i} = 4$$

So we have a sum of cube roots of complex numbers which nevertheless happens to produce a real result. So I presume that these two cube roots must be complex conjugates of each other, which seems to be the case, judging by the fact that the numbers under the cube roots are complex conjugates of each other as well (note the signs I marked with red).
Complex conjugates are "mirror images" of each other, so when added up, they produce a real result.
Cube roots of complex conjugates divide their angles by 3, so the results should remain complex conjugates, and I suppose this is the reason why they add up to a real number as well. Am I right?

What bothers me, though, is how can we PROVE that identity with algebra?

Here's what I tried:
I cubed the equation:

$$x \;=\; \sqrt[3]{2 {\color{red}+} 11i} + \sqrt[3]{2 {\color{red}-} 11i} \;=\; 4\\ x^3 \;=\; \left( \sqrt[3]{2 {\color{red}+} 11i} + \sqrt[3]{2 {\color{red}-} 11i} \right)^3 \;=\; 4^3 \;=\; 64$$

expanded the middle one from the binomial theorem:

$$x^3 \;=\; \left(\sqrt[3]{2 {\color{red}+} 11i}\right)^3 + \left(\sqrt[3]{2 {\color{red}-} 11i}\right)^3 + 3\!\cdot\!\sqrt[3]{2 {\color{red}+} 11i}\!\cdot\!\sqrt[3]{2 {\color{red}-} 11i}\!\cdot\!\left(\sqrt[3]{2 {\color{red}+} 11i} + \sqrt[3]{2 {\color{red}-} 11i}\right) \;=\; 4^3 \;=\; 64\\ x^3 \;=\; 2 {\color{red}+} 11i + 2 {\color{red}-} 11i + 3\!\cdot\!\sqrt[3]{\left(2 {\color{red}+} 11i\right)\left(2 {\color{red}-} 11i\right)}\!\cdot\!\left(\sqrt[3]{2 {\color{red}+} 11i} + \sqrt[3]{2 {\color{red}-} 11i}\right) \;=\; 4^3 \;=\; 64\\ x^3 \;=\; 4 + 3\!\cdot\!\sqrt[3]{2^2 - (11i)^2}\!\cdot\!\left(\sqrt[3]{2 {\color{red}+} 11i} + \sqrt[3]{2 {\color{red}-} 11i}\right) \;=\; 4^3 \;=\; 64\\ x^3 \;=\; 4 + 3\!\cdot\!\sqrt[3]{4 + 121}\!\cdot\!\left(\sqrt[3]{2 {\color{red}+} 11i} + \sqrt[3]{2 {\color{red}-} 11i}\right) \;=\; 4^3 \;=\; 64\\ x^3 \;=\; 4 + 3\!\cdot\!\sqrt[3]{125}\!\cdot\!\left(\sqrt[3]{2 {\color{red}+} 11i} + \sqrt[3]{2 {\color{red}-} 11i}\right) \;=\; 4^3 \;=\; 64\\ x^3 \;=\; 4 + 3\!\cdot\!5\!\cdot\!\left(\sqrt[3]{2 {\color{red}+} 11i} + \sqrt[3]{2 {\color{red}-} 11i}\right) \;=\; 4^3 \;=\; 64\\ x^3 \;=\; 4 + 15\!\cdot\!\left(\sqrt[3]{2 {\color{red}+} 11i} + \sqrt[3]{2 {\color{red}-} 11i}\right) \;=\; 4^3 \;=\; 64$$

But now the expression that remained in parentheses is just the original $x$ I started with! When trying to find the answer, I run across the original question again :/ Not only that, but replacing it with $x$ gives me the original cubic equation I started with ;/

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

Why am I going in circles with this? And what other techniques can I use to crack this and prove that this sum of cube roots indeed equals to the real number $4$?

Inb4: I already ascertained geometrically that this sum of cubes is really equal to $4$, but now I'd like to have it proved algebraically, and learn a general method of dealing with such sums of cubes of complex conjugates.

Edit: All the answers so far seem to be based on the assumption that I know that this complex expression is equal to 2 already (e.g. by restoring the original cubic equation and finding its rational roots). What I'm rather interested in, is how to find the equivalent real solutions when restoring the original cubic equation doesn't work, because it cannot be solved by the rational roots theorem.

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    $\begingroup$ It is possible for a cubic to be expressible in terms of radicals, but not to be expressible in terms of real radicals (see, e.g., this sci.math post). In fact, this is true of any cubic with three real irrational roots. That is, there is no general algebraic way to eliminate the complex numbers inside an expression of the form $\sqrt[3]{a+bi}+\sqrt[3]{a-bi}$, just the possibility of getting lucky in specific cases. $\endgroup$ – Micah Feb 25 '18 at 18:11
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    $\begingroup$ For background information: en.wikipedia.org/wiki/Casus_irreducibilis . Seach for cubic casus irreducibilis for even more, and history. $\endgroup$ – Ethan Bolker Feb 25 '18 at 18:26
  • $\begingroup$ Is there a particular format of nonrational answer you are looking for a solution to? (e.g. $a+\sqrt{b}+\sqrt[3]{c}$ for rational $a,b,c$) That might help people see if there is a method could still work for a nonrational case. $\endgroup$ – Mark S. Feb 26 '18 at 12:21
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Just note that $(2+i)^3=2+11i$ and that $(2-i)^3=2-11i$. So, a natural choice is to do\begin{align}\sqrt[3]{2+11i}+\sqrt[3]{2-11i}&=2+i+2-i\\&=4.\end{align}

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    $\begingroup$ Yeah, it's cool when you know the right solution upfront. But what if you don't? If I gave you some other, less known sum of cube roots, I guess you will be none the wiser, because then you would have to guess the right number to be cubed to get the result. And this is exactly the root of the problem: how to calculate that number algebraically instead of guessing or telling the right answer from history. $\endgroup$ – BarbaraKwarc Feb 25 '18 at 11:24
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    $\begingroup$ It's not so much of a guess. The norm of $2+11i$ is $125$, so you need a number with norm $5$. And $(2+i)$ fits the bill nicely. You can always write the number in polar form and see that the $i\sin \theta$'s cancel. $\endgroup$ – B. Goddard Feb 25 '18 at 12:40
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Let's recall the derivation of the solution of the cubic equation $x^3+px+q=0$: We assume $x=u+v$, and get $$x^3+px+q=u^3+v^3+q+(3uv+p)(u+v).$$ So if we additionally assume $$3uv+p=0\tag1,$$ we must also have $$u^3+v^3+q=0\tag2,$$ and from (1) and (2) we easily get a quadratic equation for $u^3, v^3$. So $x$ is a sum of two cube roots, but those aren't independent, they must satisfy (1). In the complex case, there are 3 values for each cube root, giving 9 combinations, but (1) reduces that to 3 valid combinations. In your case, the other two valid combinations give the other two solutions of your cubic equation, both are real, too: $x=-2\pm\sqrt{3}$.
You get those from the quadratic equation $x^2+4x+1=0$, resulting from division of $x^3-15x-4$ by $x-4$.

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  • $\begingroup$ I know, I solved the cubic already, both ways. That's how I know that this sum of cube roots is also equal to 2. But if I gave you only the sum of the cube roots (not necessarily this particular one), how would you find out their sum then? :q Sure, you can restore the original equation (it comes out when you cube both sides, trying to simplify this sum of cube roots, as I showed), but then we're going in circles again. $\endgroup$ – BarbaraKwarc Feb 25 '18 at 12:16
  • $\begingroup$ In this particular case, it is easy to find that this expression has to be equal to 2, by trying out the divisors of 4 in the restored cubic equation. But you won't be always so lucky. Try this one, for example: $x = \sqrt[3]{\frac{-1+i\sqrt{3}}{16}} + \sqrt[3]{\frac{-1-i\sqrt{3}}{16}}$. This number is real as well, but restoring the original cubic equation and solving it won't help you much to find it. $\endgroup$ – BarbaraKwarc Feb 25 '18 at 12:20
  • $\begingroup$ @BarbaraKwarc Once again, not both of us move in circles: you give an expression having 9 different interpretations. Three of them are pairs of conjugates, so their sum will be real ($\cos\dfrac{2\pi}9, \cos\dfrac{8\pi}9, \cos\dfrac{14\pi}9$, I'd say). So which of those three values did you mean? $\endgroup$ – Professor Vector Feb 25 '18 at 12:37
  • $\begingroup$ In the original question, I meant the $4$. In this particular question, I would have one of those cosines which the expression should be equal to (let's say the first cosine). But no matter which one you choose, the point still stands: you will be moving in circles when you try simplifying it by cubing both sides. You'll end up with the original cubic equation, which in this case won't have any rational roots, and the values for the cosines can only be approximated as far as I know, so I wouldn't call it a solution, rather a restatement of the problem in a different notation. $\endgroup$ – BarbaraKwarc Feb 25 '18 at 13:04
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This won't always work out so nicely, and there is likely a shortcut or at least a more efficient way of solving the geometric equations, but here is one approach that works out for a variety of special cases, including $\sqrt[3]{2+11i}+\sqrt[3]{2-11i}$.

Using the polar form of complex numbers, we can see that the value we're looking for is $2\sqrt{5}\cos\left(\dfrac{\arctan\frac{11}{2}}{3}\right)$. We can figure out this cosine by drawing the triangle in the complex plane with vertices $0$, $2$, and $2+11i$ and trisecting the angle at $0$.

right triangle with trisected angle

Then we have the following equations:

  1. $a+b+c=11$ because that is the height of the triangle.
  2. $2^2+a^2=x^2$ by Pythagorean Theorem on smallest triangle.
  3. $2^2+(a+b)^2=y^2$ by Pythagorean Theorem.
  4. $2^2+11^2=z^2$ by Pythagorean Theorem.
  5. $\frac{b}{a}=\frac{y}{2}$ by Angle Bisector Theorem.
  6. $\frac{c}{b}=\frac{z}{x}$ by Angle Bisector Theorem.
  7. $\frac{2}{x}$ is the cosine we're looking for by right triangle trig.

Now we just have to do the terrible algebra: We can eliminate $z$ and $a$ with equations 6 and 2 respectively, to get:

  1. $\frac{b}{\sqrt{x^2-4}}=\frac{y}{2}$ (from 5 above)
  2. $125b^2=x^2c^2$ (from 4 above)
  3. $4+\left(\sqrt{x^2-4}+b\right)^2=y^2$ (from 3 above)
  4. $\sqrt{x^2-4}+b+c=11$ (from 1 above)

Then we can eliminate $y$ with equation 1 to get:

  1. $125b^2=x^2c^2$ (2 above)
  2. $4+\left(\sqrt{x^2-4}+b\right)^2=\dfrac{4b^2}{x^2-4}$ (from 3 above)
  3. $\sqrt{x^2-4}+b+c=11$ (4 above)

From here we have a few options that all can work out (we can eliminate $x$, for instance). Since $x$ is our goal, here is an approach headed that way: Equation 1 tells us that $c=5\sqrt 5b/x$ and equation 3 can be used to eliminate the square root from equation 2, which leaves us with:

  1. $4+(11-5\sqrt5\frac{b}x)^2=\frac{4b^2}{x^2-4}$
  2. $\sqrt{x^2-4}+b+5\sqrt{5}\frac{b}x=11$

Solving 1. for $b$ yields $b=5\sqrt{5}x\dfrac{11x^2-44\pm4\sqrt{x^2-4}}{121x^2-500}$. Since we know $x>2$ and $b<x$, we need the negative root here: $b=5\sqrt{5}x\dfrac{11x^2-44-4\sqrt{x^2-4}}{121x^2-500}$.

Solving 2. for $b$ yields $b=\dfrac{x(11-\sqrt{x^4-2})}{x+5\sqrt{5}}$.

Setting these equal gives us a complicated expression that we can manipulate to solve for the square root: $$\sqrt{x^2-4}=\dfrac{11\left(5\sqrt{5}x^3+4x^2-20\sqrt{5}x\right)}{-121x^2+20\sqrt{5}x+1000}$$

Squaring both sides and putting over a common denominator (and dividing by $4$) gives us: $$\dfrac{121x^6+2420\sqrt{5}x^5+44875x^4-19680\sqrt{5}x^3-429500x^2+40000\sqrt{5}x+1000000}{\left(-121x^2+20\sqrt{5}x+1000\right)^2}=0$$

We only have to worry about the numerator, and since we have $\sqrt{5}$ as a factor of every other coefficient, it will be convenient to substitute $x=\hat{x}*\sqrt{5}$, so that the equation from the numerator becomes $15125\hat{x}^6+302500\hat{x}^5+1121875\hat{x}^4-492000\hat{x}^3-2147500\hat{x}^2-200000\hat{x}+1000000=0$. Testing rational roots tells us we have $\hat{x}=-5,1,-\frac{10}{11},\frac{10}{11}$ as roots. The remaining ones are roots of the resulting quadratic: $\hat{x}=4(-2\pm\sqrt{3})$.

We know $\hat{x}$ must be positive, so it's either $1$ or $10/11$, and checking the other equations shows that $\hat{x}=1$ is correct, which leads to the cosine being $\dfrac{2}{\sqrt{5}*1}$ and the original value being $2\sqrt{5}*\dfrac{2}{\sqrt{5}*1}=4\checkmark$.

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  • $\begingroup$ This was my first idea, I didn't check all step but it seems a very nice work and it seems very difficult to obtain the result! (+1) $\endgroup$ – user Feb 26 '18 at 7:13
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As an alternative we could show algebrically that

$$x=z+\bar z\implies x=2Re(z)=2\sqrt 5\cdot \left(\cos \frac{\theta}3\right)=4$$

with

$$\theta = \arccos \frac{2}{5\sqrt5}$$

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  • $\begingroup$ That's nonsense, sorry. $\endgroup$ – Professor Vector Feb 25 '18 at 12:31
  • $\begingroup$ Why Is it nonsense? $\endgroup$ – user Feb 25 '18 at 12:34
  • $\begingroup$ It doesn't help the OP to make the same mistake (arbitrarily choosing one of three possible roots), especially if the connection to the original question is so obscure as in you mass-produced "answer". $\endgroup$ – Professor Vector Feb 25 '18 at 12:53
  • $\begingroup$ @ProfessorVector yeah, the notation kinda is, but I understand what he meant to say, because this is the "geometric way" I mentioned in my question. Complex conjugates are symmetric with respect to the real axis, so when we add them, they form the sides of a parallelogram, with its diameter laying along the real axis. Half that sum is the perpendicular projection of each of the conjugates onto the real axis, that is, its real component. So the sum is twice the real component of the conjugate. He used that, and the fact that the cube root divides the angle by 3, hence $\cos(\frac{\theta}{3})$ $\endgroup$ – BarbaraKwarc Feb 25 '18 at 12:53
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    $\begingroup$ That's the thing: this particular equation is well-known from historic reasons, and everyone knows what the solution is, so everyone is jumping to conclusions and tells the solution from memory. But if I chose a different pair of cube roots, less known, which are the solution to a cubic equation that doesn't have rational roots (but that are still real and algebraic), I bet people would be none the wiser and guessing or reciting solutions from memory won't quite work. The discussion about other values of cube roots is a moot point, only to muddle the real issue here. $\endgroup$ – BarbaraKwarc Feb 25 '18 at 13:08
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Here's an odd approach: the rational root theorem.

Possible factors: $\pm 1,\pm 2,\pm 4$

Using trial and error along with synthetic division upon this polynomial shows that $4$ is a solution.

This doesn't solve your question on how to prove the idendity, but it does provide a simple solution to similar questions that will have this method easily applicable.

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The goal is to show that

$$\sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i} = 4$$

I think you're going to have problems proving that algebraically.

From the standpoint of an unordered field, where the only operations available are $+ - \times\ \div$, it's impossible, since the three roots of each number are indistinguishable from each other, and for some pairings of those values, the identity is false.

So, we have to work in the complex plane. The essence of the problem is to find the real component of $\sqrt[3]{2+11i}$. The "obvious" way to do that is to compute an arctangent, divide the angle by three, then take the cosine. Actually carrying out this computation, in the general case, and without approximation, is beyond the ability of our current mathematics.

It's an angle trisection problem, and that means that you're always going to come back to solving a cubic equation.

I'd suggest approaching the factorization problem differently, using Kronecker's algorithm (older but easier to understand), or Cantor-Zassenhaus (most widely implemented in computer algebra systems). See wikipedia's article on Polynomial Factorization.

The idea is to factor the polynomial over the rational number field, not the complex field. If your answer isn't rational, this won't work (the algorithm will tell you that the polynomial is irreducible), but if one of your roots is a rational number (like 4), this method will find it without any circular arguments.

What if your answer isn't rational? Then the best you can hope for is to find an irreducible polynomial with a single real root that is your solution, and I think you're running up against fundamental theoretical limits to do any better.

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I think all your way is total wrong because $\sqrt[3]{2+11i}$ is a set of three numbers and $\sqrt[3]{2-11i}$ is a set of three numbers.

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  • $\begingroup$ I'm aware of that, but since I'm looking for a real solution, I may safely assume that taking only the real answer (i.e. the principal root) would bring me closer to that real solution. $\endgroup$ – BarbaraKwarc Feb 25 '18 at 11:26
  • $\begingroup$ It's very common for the cubic formula to be written in this sort of way, since you can get to the answers to the original equation regardless of your choice of cube root. In this case, BarbaraKwarc has already said that they were interested in the choice that makes the sum 4. $\endgroup$ – Mark S. Feb 26 '18 at 3:02

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