已知在 $\left( x - \dfrac{k}{x^{3}} \right)^{8}$ 的二项式展开式中，常数项为 7，求正常数 $k$ 的值。 利用在第 (a) 问中求得的 $k$ 值，证明在 $\left(1 + x^{4}\right)\left( x - \dfrac{k}{x^{3}} \right)^{8}$ 的展开式中没有常数项。

1. Write down the general term of the binomial expansion

The general term is given by:
$T_{r+1} = \binom{8}{r} (x)^{8-r} \left(-\frac{k}{x^3}\right)^r$
$T_{r+1} = \binom{8}{r} (-k)^r x^{8-r} x^{-3r}$
$T_{r+1} = \binom{8}{r} (-k)^r x^{8-4r}$

2. Find the value of $r$ for the constant term

For the constant term, the power of $x$ must be 0:
$8 - 4r = 0$
$4r = 8$
$r = 2$

3. Equate the coefficient of the constant term to 7

Substitute $r = 2$ into the coefficient part:
$\binom{8}{2} (-k)^2 = 7$
$28k^2 = 7$

4. Solve for $k$ and select the positive value

$k^2 = \frac{7}{28} = \frac{1}{4}$
$k = \pm\frac{1}{2}$
Since $k$ is a positive constant, $k = \frac{1}{2}$.

1. Find the value of $k$ from part (a)

The general term of $\left( x - \frac{k}{x^{3}} \right)^{8}$ is $T_{r+1} = \binom{8}{r} x^{8-r} \left( -\frac{k}{x^{3}} \right)^{r} = \binom{8}{r} (-k)^{r} x^{8-4r}$.
For the constant term, $8 - 4r = 0 \implies r = 2$.
The constant term is $\binom{8}{2} (-k)^{2} = 28k^{2}$.
Given $28k^{2} = 7 \implies k^{2} = \frac{1}{4}$. Since $k > 0$, $k = \frac{1}{2}$.

2. Set up the expression for the constant term in the new expansion

We need the constant term in $\left(1 + x^{4}\right)\left( x - \frac{1/2}{x^{3}} \right)^{8}$.
This is formed by:
$1 \times \left(\text{constant term of } \left( x - \frac{1/2}{x^{3}} \right)^{8}\right) + x^{4} \times \left(\text{term in } x^{-4} \text{ of } \left( x - \frac{1/2}{x^{3}} \right)^{8}\right)$.

3. Find the term in $x^{-4}$ of the binomial expansion

For the term in $x^{-4}$, we set the power of $x$ in the general term to $-4$:
$8 - 4r = -4 \implies 4r = 12 \implies r = 3$.
The coefficient of $x^{-4}$ is $\binom{8}{3} \left(-\frac{1}{2}\right)^{3} = 56 \left(-\frac{1}{8}\right) = -7$.

4. Calculate the overall constant term to show it is zero

The constant term of $\left( x - \frac{1/2}{x^{3}} \right)^{8}$ is 7 (given in part a).
The overall constant term is $1(7) + 1(-7) = 7 - 7 = 0$.
Therefore, there is no constant term in the expansion.