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.
Best Answer
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}