在 $(2x-1)^2\left(1+\dfrac{p}{x}\right)^8$ 的展开式中（其中 $p$ 是正常数），不含 $\dfrac{1}{x^3}$ 项。求常数 $p$ 的可能值。

1. Expand the first bracket.

$(2x-1)^2 = 4x^2 - 4x + 1$

2. Write down the general term for the second bracket.

The general term of $\left(1+\dfrac{p}{x}\right)^8$ is $T_{r+1} = \binom{8}{r} (1)^{8-r} \left(\dfrac{p}{x}\right)^r = \binom{8}{r} p^r x^{-r}$.

3. Identify the terms in the second bracket that will multiply with the first bracket to give $x^{-3}$.

To obtain the term in $x^{-3}$ (or $\dfrac{1}{x^3}$), we need:

• $4x^2$ multiplied by the term in $x^{-5}$ ($r=5$)
• $-4x$ multiplied by the term in $x^{-4}$ ($r=4$)
• $1$ multiplied by the term in $x^{-3}$ ($r=3$)

4. Form an equation for the coefficient of $x^{-3}$.

Coefficient of $x^{-3} = 4\binom{8}{5}p^5 - 4\binom{8}{4}p^4 + 1\binom{8}{3}p^3$
$= 4(56)p^5 - 4(70)p^4 + 56p^3$
$= 224p^5 - 280p^4 + 56p^3$
Since there is no term in $x^{-3}$, this coefficient is $0$:
$224p^5 - 280p^4 + 56p^3 = 0$

5. Solve the equation for $p$.

Since $p$ is a positive constant, $p \neq 0$. Divide by $56p^3$:
$4p^2 - 5p + 1 = 0$
$(4p - 1)(p - 1) = 0$
$p = \frac{1}{4}$ or $p = 1$