# How do you use the Binomial Theorem to find the value of 999^3?

May 28, 2016

Rewrite ${999}^{3}$ as ${\left({10}^{3} - 1\right)}^{3}$ and apply the binomial theorem to find that

${999}^{3} = 997002999$

#### Explanation:

The binomial theorem states that ${\left(a + b\right)}^{n} = {\sum}_{k = 0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) {a}^{n - k} {b}^{k}$

where ((n),(k)) = (n!)/(k!(n-k)!)

For this problem, we will only need to calculate for $n = 3$, and we will find that $\left(\begin{matrix}3 \\ 0\end{matrix}\right) = \left(\begin{matrix}3 \\ 3\end{matrix}\right) = 1$ and $\left(\begin{matrix}3 \\ 1\end{matrix}\right) = \left(\begin{matrix}3 \\ 2\end{matrix}\right) = 3$
(Try verifying this)

Noting that it is much easier to calculate powers of $10$ and $1$ compared to $999$, we can rewrite $999$ as $1000 - 1 = {10}^{3} - 1$ and apply the binomial theorem:

${999}^{3} = {\left({10}^{3} - 1\right)}^{3}$

$= \left(\begin{matrix}3 \\ 0\end{matrix}\right) {\left({10}^{3}\right)}^{3} + \left(\begin{matrix}3 \\ 1\end{matrix}\right) {\left({10}^{3}\right)}^{2} \left(- 1\right) + \left(\begin{matrix}3 \\ 2\end{matrix}\right) \left({10}^{3}\right) {\left(- 1\right)}^{2} + \left(\begin{matrix}3 \\ 3\end{matrix}\right) {\left(- 1\right)}^{3}$

$= {10}^{9} - 3 \cdot {10}^{6} + 3 \cdot {10}^{3} - 1$

$= 1000000000 - 3000000 + 3000 - 1$

$= 997002999$