Given
\begin{align}
a &= {2015\choose0}+{2015\choose3}+{2015\choose6}+\cdots \\[2pt]
b &= {2015\choose1}+{2015\choose4}+{2015\choose7}+\cdots \\[2pt]
c &= {2015\choose2}+{2015\choose5}+{2015\choose8}+\cdots
\end{align}
We have to find $(b-c)^2+(c-a)^2$
Note: Here $${n\choose{k}}=\frac{n!}{k!\cdot(n-k)!}$$ where
$k \leq n$ for all $n,k \in \mathbb{N}$. It is called the binomial coefficient.
I have observed that $(a+b+c)=2^{2015}$ using the binomial expansion but the given fact does not provide clue to solve the given problem. Hence how can we solve the given problem?
Best Answer
This is along the general lines of Feng's comment and John's post, with just simpler algebra.
As you noted, $a+b+c=(1+1)^{2015} = 2^{2015}=n$, say. Using the non-real cube root of unity which satisfies $\omega^3=1$ and $1+\omega+\omega^2=0$, you can similarly note that
$a+b\omega+c\omega^2=(1+\omega)^{2015}=(-\omega^2)^{2015}=-\omega$ and $a+b\omega^2+c\omega=(1+\omega^2)^{2015}=(-\omega)^{2015}=-\omega^2$.
Adding those three, we get
$3a=n-\omega-\omega^2=n+1$. Similarly using the above three equations, we also get
$3b=n+(-\omega)\omega^2+(-\omega^2)\omega=n-2$ and
$3c=n+(-\omega)\omega+(-\omega^2)\omega^2=n-\omega^2-\omega=n+1$
Hence $3b-3c=-3$ and $3c-3a=0$, giving $(b-c)^2+(c-a)^2=(-1)^2+0=1.$