Expand $$$\left(x + 5\right)^{8}$$$

Expand $$$\left(x + 5\right)^{8}$$$.

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 $$$a = x$$$, $$$b = 5$$$, and $$$n = 8$$$.

Therefore, $$$\left(x + 5\right)^{8} = \sum_{k=0}^{8} {\binom{8}{k}} x^{8 - k} 5^{k}$$$.

Now, calculate the product for every value of $$$k$$$ from $$$0$$$ to $$$8$$$.

$$$k = 0$$$: $$${\binom{8}{0}} x^{8 - 0} \cdot 5^{0} = \frac{8!}{\left(8 - 0\right)! 0!} x^{8 - 0} \cdot 5^{0} = x^{8}$$$

$$$k = 1$$$: $$${\binom{8}{1}} x^{8 - 1} \cdot 5^{1} = \frac{8!}{\left(8 - 1\right)! 1!} x^{8 - 1} \cdot 5^{1} = 40 x^{7}$$$

$$$k = 2$$$: $$${\binom{8}{2}} x^{8 - 2} \cdot 5^{2} = \frac{8!}{\left(8 - 2\right)! 2!} x^{8 - 2} \cdot 5^{2} = 700 x^{6}$$$

$$$k = 3$$$: $$${\binom{8}{3}} x^{8 - 3} \cdot 5^{3} = \frac{8!}{\left(8 - 3\right)! 3!} x^{8 - 3} \cdot 5^{3} = 7000 x^{5}$$$

$$$k = 4$$$: $$${\binom{8}{4}} x^{8 - 4} \cdot 5^{4} = \frac{8!}{\left(8 - 4\right)! 4!} x^{8 - 4} \cdot 5^{4} = 43750 x^{4}$$$

$$$k = 5$$$: $$${\binom{8}{5}} x^{8 - 5} \cdot 5^{5} = \frac{8!}{\left(8 - 5\right)! 5!} x^{8 - 5} \cdot 5^{5} = 175000 x^{3}$$$

$$$k = 6$$$: $$${\binom{8}{6}} x^{8 - 6} \cdot 5^{6} = \frac{8!}{\left(8 - 6\right)! 6!} x^{8 - 6} \cdot 5^{6} = 437500 x^{2}$$$

$$$k = 7$$$: $$${\binom{8}{7}} x^{8 - 7} \cdot 5^{7} = \frac{8!}{\left(8 - 7\right)! 7!} x^{8 - 7} \cdot 5^{7} = 625000 x$$$

$$$k = 8$$$: $$${\binom{8}{8}} x^{8 - 8} \cdot 5^{8} = \frac{8!}{\left(8 - 8\right)! 8!} x^{8 - 8} \cdot 5^{8} = 390625$$$

Thus, $$$\left(x + 5\right)^{8} = x^{8} + 40 x^{7} + 700 x^{6} + 7000 x^{5} + 43750 x^{4} + 175000 x^{3} + 437500 x^{2} + 625000 x + 390625.$$$

$$$\left(x + 5\right)^{8} = x^{8} + 40 x^{7} + 700 x^{6} + 7000 x^{5} + 43750 x^{4} + 175000 x^{3} + 437500 x^{2} + 625000 x + 390625$$$A