What is the easiest way to get: $2+ \sqrt{-121} = (2+ \sqrt{-1})^3$ I was reading the book Seventeen equations have changed the world.
At some point, while the book was talking about complex numbers, I see this equation:
$2+ \sqrt{-121} = (2+ \sqrt{-1})^3$
Even if it's easy to proof the truth of this equivalence (it is enough to develop the two members),
I can't find an easy/good/fast way to obtain straight the identity. 
Can you help me? Does there exist a mathematical property that I'm missing?
 A: Expanding the product as
\begin{eqnarray*}
(2+\sqrt{-1})^3&=&2^3+3\times2^2\sqrt{-1}+3\times2\sqrt{-1}^2+\sqrt{-1}^3\\
&=&(2^3-6)+(3\times2^2-1)\sqrt{-1}\\
&=&2+11\sqrt{-1}\\
&=&2+\sqrt{-121},
\end{eqnarray*}
is so elementary that I cannot think of any easier way to show it.
But suppose you want to determine the factorization of $2+\sqrt{-121}$, i.e. roughly speaking an expression
$$2+\sqrt{-121}=(a_1+b_1\sqrt{-1})(a_2+b_2\sqrt{-1})\cdots(a_n+b_n\sqrt{-1}),$$
where the $a_i$ and $b_i$ are integers. This is the subject of algebraic number theory. I will use a few results without mention, as I don't know your background in the subject. You could note that the norm equals
$$N(2+\sqrt{-121})=N(2+11\sqrt{-1})=2^2+11^2=125=5^3,$$
and that $5$ does not divide $2+\sqrt{-121}$. This implies that
$$2+11\sqrt{-1}=(a+b\sqrt{-1})^3,$$
where the norm of $a+b\sqrt{-1}$ equals $5$. This means that
$$5=N(a+b\sqrt{-1})=a^2+b^2,$$
which has the solutions
$$1+2\sqrt{-1},\qquad 1-2\sqrt{-1},\qquad 2+\sqrt{-1},\qquad 2-\sqrt{-1}.$$
It is then a quick check to show that only $2+\sqrt{-1}$ works.
A: Not sure how you approached it before.
If you want to utlise the binomial expansion
$$
(x+y)^n = \sum_{k=0}^n\left(\matrix{n\\k}\right)x^ky^{n-k}
$$
then we have
$$
\begin{align}
(x+y)^3 &= \left(\matrix{3\\0}\right)2^3 +\left(\matrix{3\\1}\right)2^2\sqrt{-1} + \left(\matrix{3\\2}\right)2^1(\sqrt{-1})^2  + \left(\matrix{3\\3}\right)(\sqrt{-1})^3 \\
&= 2^3 + 3\cdot 2^2\sqrt{-1} + 3\cdot 2\cdot-1 + \sqrt{-1}\cdot(\sqrt{-1})^2\\
&=8+12\sqrt{-1} -6 -\sqrt{-1} = 2 +11\sqrt{-1} = 2 + \sqrt{11^2\cdot - 1} \\
&=2+\sqrt{-121}
\end{align} 
$$
A: The development is easy.
Mentally,
$$(2+i)^3=8-6+i(12-1).$$
I can't see any shortcut. (Of course, the identity in the book was established from the right.)
A: If it is to prove:
$$2+11i=2+i+10i=2+i+(2+i)(2+4i)=(2+i)(3+4i)=(2+i)(2+i+1+3i)=(2+i)(2+i+(2+i)(1+i))=(2+i)(2+i)(2+i).$$
