Expand $$$\left(a + b\right)^{4}$$$

Expand $$$\left(a + b\right)^{4}$$$.

The expansion is given by the following formula: $$$\left(a + b\right)^{n} = \sum_{k=0}^{n} {\binom{n}{k}} a^{n - k} b^{k}$$$, where $$${\binom{n}{k}} = \frac{n!}{\left(n - k\right)! k!}$$$ and $$$n! = 1 \cdot 2 \cdot \ldots \cdot n$$$.

We have that $$$a = a$$$, $$$b = b$$$, and $$$n = 4$$$.

Therefore, $$$\left(a + b\right)^{4} = \sum_{k=0}^{4} {\binom{4}{k}} a^{4 - k} b^{k}$$$.

Now, calculate the product for every value of $$$k$$$ from $$$0$$$ to $$$4$$$.

$$$k = 0$$$: $$${\binom{4}{0}} a^{4 - 0} b^{0} = \frac{4!}{\left(4 - 0\right)! 0!} a^{4 - 0} b^{0} = a^{4}$$$

$$$k = 1$$$: $$${\binom{4}{1}} a^{4 - 1} b^{1} = \frac{4!}{\left(4 - 1\right)! 1!} a^{4 - 1} b^{1} = 4 a^{3} b$$$

$$$k = 2$$$: $$${\binom{4}{2}} a^{4 - 2} b^{2} = \frac{4!}{\left(4 - 2\right)! 2!} a^{4 - 2} b^{2} = 6 a^{2} b^{2}$$$

$$$k = 3$$$: $$${\binom{4}{3}} a^{4 - 3} b^{3} = \frac{4!}{\left(4 - 3\right)! 3!} a^{4 - 3} b^{3} = 4 a b^{3}$$$

$$$k = 4$$$: $$${\binom{4}{4}} a^{4 - 4} b^{4} = \frac{4!}{\left(4 - 4\right)! 4!} a^{4 - 4} b^{4} = b^{4}$$$

Thus, $$$\left(a + b\right)^{4} = a^{4} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + b^{4}$$$.

$$$\left(a + b\right)^{4} = a^{4} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + b^{4}$$$A