写出 $\left(x^{2}+\dfrac{m}{x}\right)^{9}$ 按 x 的降幂展开式中的前三项，其中 $m$ 是一个整数。 已知在 $\left(x^{2}+\dfrac{m}{x}\right)^{9}$ 的展开式中，$x^{3}$ 的系数为 $-126$， 证明 $m=-1$。 由此，求 $\left(2-\dfrac{1}{x^{3}}\right)\left(x^{2}+\dfrac{m}{x}\right)^{9}$ 的展开式中的常数项。

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

$\left(x^{2}+\frac{m}{x}\right)^{9} = \binom{9}{0}(x^{2})^{9} + \binom{9}{1}(x^{2})^{8}\left(\frac{m}{x}\right) + \binom{9}{2}(x^{2})^{7}\left(\frac{m}{x}\right)^{2} + \dots$

2. Simplify the terms to obtain the final expression in descending powers of x.

$= 1(x^{18}) + 9(x^{16})\left(\frac{m}{x}\right) + 36(x^{14})\left(\frac{m^{2}}{x^{2}}\right) + \dots$
$= x^{18} + 9mx^{15} + 36m^{2}x^{12}$

1. Write down the general term for the expansion of $\left(x^{2}+\frac{m}{x}\right)^{9}$

The general term is given by:
$T_{r+1} = \binom{9}{r} (x^2)^{9-r} \left(\frac{m}{x}\right)^r$
$T_{r+1} = \binom{9}{r} x^{18-2r} \cdot m^r x^{-r}$
$T_{r+1} = \binom{9}{r} m^r x^{18-3r}$

2. Find the value of $r$ for the term in $x^3$

For the term in $x^3$, we equate the power of $x$ to 3:
$18 - 3r = 3$
$3r = 15$
$r = 5$

3. Equate the coefficient to $-126$ and solve for $m$

Substitute $r = 5$ into the coefficient part of the general term:
$\text{Coefficient} = \binom{9}{5} m^5$
$126 m^5 = -126$
$m^5 = -1$
$m = -1 \quad \text{(shown)}$