Awesome Binomial Theorem!

Let $S= \sum_{r=0}^n {n \choose r} b^{(n-r)} c^{r} \sin \ {rB-(n-r)C}$ and $T= \sum_{r=0}^n {n \choose r} b^{(n-r)} c^{r} \cos \ {rB-(n-r)C}$ Then, $T+iS= \sum_{r=0}^n {n \choose r} b^{(n-r)} c^{r} e^{i [rB-(n-r)C]}$ $= \sum_{r=0}^n {n \choose r} (b e^{-iC})^{(n-r)} (c e^{iB})^{r}$ $= (b e^{-iC} + c e^{iB})^{n}$ $= [(b \cos \ C + c \cos \ B) + i(-b \sin \ C + c \sin \ B)]^{n}$ But in any triangle, $b \cos \ C + c \cos \ B = a$ and $\frac {b}{\sin \ B} = \frac {c}{\sin \ C}$ Thus $T+iS = a^{n}$

And hence $T = a^{n}$ and $S = 0$ and $N = {10 \choose 5} = 252$
Therefore, $N-S = 252$