写出并化简 $\left(2 - \dfrac{3}{x}\right)^8$ 按 $x$ 的降幂排列的展开式中的前三项。 已知在 $(1 - 2x - kx^2)\left(2 - \dfrac{3}{x}\right)^8$ 的展开式中没有 $x$ 项，求该展开式中的常数项。

1. Apply the binomial theorem to write the first three terms

$\left(2 - \frac{3}{x}\right)^8 = 2^8 + \binom{8}{1}2^7\left(-\frac{3}{x}\right) + \binom{8}{2}2^6\left(-\frac{3}{x}\right)^2 + \dots$

2. Simplify the terms

$= 256 + 8(128)\left(-\frac{3}{x}\right) + 28(64)\left(\frac{9}{x^2}\right) + \dots$
$= 256 - \frac{3072}{x} + \frac{16128}{x^2}$

1. Find the first three terms of the expansion of $\left(2 - \frac{3}{x}\right)^8$ from part (a).

The first three terms in descending powers of $x$ are:
$\binom{8}{0}2^8 + \binom{8}{1}2^7\left(-\frac{3}{x}\right) + \binom{8}{2}2^6\left(-\frac{3}{x}\right)^2$
$= 256 - \frac{3072}{x} + \frac{16128}{x^2}$

2. Form an equation for the coefficient of $x$ in the product and solve for $k$.

The expansion is $(1 - 2x - kx^2)\left(256 - \frac{3072}{x} + \frac{16128}{x^2} + \dots\right)$.
The $x$ term is formed by:
$(-2x)(256) + (-kx^2)\left(-\frac{3072}{x}\right) = -512x + 3072kx$
Given there is no $x$ term:
$-512 + 3072k = 0$
$3072k = 512 \implies k = \frac{1}{6}$

3. Calculate the constant term using the value of $k$.

The constant term is formed by:
$(1)(256) + (-2x)\left(-\frac{3072}{x}\right) + (-kx^2)\left(\frac{16128}{x^2}\right)$
$= 256 + 6144 - 16128k$
Substitute $k = \frac{1}{6}$:
$= 6400 - 16128\left(\frac{1}{6}\right)$
$= 6400 - 2688 = 3712$