展开 $$$\left(x - 3\right)^{4}$$$

展开 $$$\left(x - 3\right)^{4}$$$。

该展开由以下公式给出：$$$\left(a + b\right)^{n} = \sum_{k=0}^{n} {\binom{n}{k}} a^{n - k} b^{k}$$$，其中$$${\binom{n}{k}} = \frac{n!}{\left(n - k\right)! k!}$$$和$$$n! = 1 \cdot 2 \cdot \ldots \cdot n$$$。

我们有$$$a = x$$$、$$$b = -3$$$和$$$n = 4$$$。

因此，$$$\left(x - 3\right)^{4} = \sum_{k=0}^{4} {\binom{4}{k}} x^{4 - k} \left(-3\right)^{k}$$$。

现在，计算 $$$k$$$ 从 $$$0$$$ 到 $$$4$$$ 的每个取值对应的乘积。

$$$k = 0$$$: $$${\binom{4}{0}} x^{4 - 0} \left(-3\right)^{0} = \frac{4!}{\left(4 - 0\right)! 0!} x^{4 - 0} \left(-3\right)^{0} = x^{4}$$$

$$$k = 1$$$: $$${\binom{4}{1}} x^{4 - 1} \left(-3\right)^{1} = \frac{4!}{\left(4 - 1\right)! 1!} x^{4 - 1} \left(-3\right)^{1} = - 12 x^{3}$$$

$$$k = 2$$$: $$${\binom{4}{2}} x^{4 - 2} \left(-3\right)^{2} = \frac{4!}{\left(4 - 2\right)! 2!} x^{4 - 2} \left(-3\right)^{2} = 54 x^{2}$$$

$$$k = 3$$$: $$${\binom{4}{3}} x^{4 - 3} \left(-3\right)^{3} = \frac{4!}{\left(4 - 3\right)! 3!} x^{4 - 3} \left(-3\right)^{3} = - 108 x$$$

$$$k = 4$$$: $$${\binom{4}{4}} x^{4 - 4} \left(-3\right)^{4} = \frac{4!}{\left(4 - 4\right)! 4!} x^{4 - 4} \left(-3\right)^{4} = 81$$$

因此，$$$\left(x - 3\right)^{4} = x^{4} - 12 x^{3} + 54 x^{2} - 108 x + 81$$$。

$$$\left(x - 3\right)^{4} = x^{4} - 12 x^{3} + 54 x^{2} - 108 x + 81$$$A