# How do you use the binomial theorem to expand and simplify the expression (a+6)^4?

Aug 27, 2017

${a}^{4} + 24 {a}^{3} + 216 {a}^{2} + 864 a + 1296$

#### Explanation:

The binomial theorem states that

${\left(a + b\right)}^{n} = \left(\begin{matrix}n \\ 0\end{matrix}\right) {a}^{n} {b}^{0} + \left(\begin{matrix}n \\ 1\end{matrix}\right) {a}^{n - 1} {b}^{1} + \left(\begin{matrix}n \\ 2\end{matrix}\right) {a}^{n - 2} {b}^{2} + \ldots + \left(\begin{matrix}n \\ n\end{matrix}\right) {a}^{0} {b}^{n}$

$\left(\begin{matrix}n \\ r\end{matrix}\right)$ is a combination read as "$n$ choose $r$".

((n),(r)) = color(white)I_nC_r = (n!) / (r! (n-r)!)

To expand ${\left(a + 6\right)}^{4}$, substitute $a = a , b = 6 ,$ and $n = 4$ into the formula.

${\left(a + 6\right)}^{4} = \left(\begin{matrix}4 \\ 0\end{matrix}\right) \cdot {a}^{4} \cdot {6}^{0} + \left(\begin{matrix}4 \\ 1\end{matrix}\right) \cdot {a}^{3} \cdot {6}^{1} + \left(\begin{matrix}4 \\ 2\end{matrix}\right) \cdot {a}^{2} \cdot {6}^{2} + \left(\begin{matrix}4 \\ 3\end{matrix}\right) \cdot {a}^{1} \cdot {6}^{3} + \left(\begin{matrix}4 \\ 4\end{matrix}\right) \cdot {a}^{0} \cdot {6}^{4}$

$= 1 \cdot {a}^{4} \cdot 1 + 4 \cdot {a}^{3} \cdot 6 + 6 \cdot {a}^{2} \cdot 36 + 4 \cdot {a}^{1} \cdot 216 + 1 \cdot 1 \cdot 1296$

$= {a}^{4} + 24 {a}^{3} + 216 {a}^{2} + 864 a + 1296$