By considering the general term in the binomial expansion of $\left(x^{4}-\frac{1}{k x^{2}}\right)^{6}$, where $k$ is a positive constant, explain why there are only even powers of $x$ in this expansion. Given that the term independent of $x$ in the binomial expansion of $\left(x^{4}-\frac{1}{k x^{2}}\right)^{6}$ is $\dfrac{5}{27}$, show that $k$ is 3. Hence find the term independent of $x$ in $\left(2-3x^{6}\right)\left(x^{4}-\dfrac{1}{k x^{2}}\right)^{6}$.

1. Write down the general term of the binomial expansion

The general term $T_{r+1}$ is given by:
$T_{r+1} = \binom{6}{r} (x^4)^{6-r} \left(-\frac{1}{kx^2}\right)^r$

2. Simplify the general term to find the power of $x$

$T_{r+1} = \binom{6}{r} x^{24-4r} \left(-\frac{1}{k}\right)^r x^{-2r}$
$T_{r+1} = \binom{6}{r} \left(-\frac{1}{k}\right)^r x^{24-6r}$

3. Explain why the power of $x$ is always even

The power of $x$ is $24 - 6r$. Since $r$ is an integer ($0 \leqslant r \leqslant 6$), $6r$ is an even integer. The difference between two even integers ($24$ and $6r$) is always an even integer. Therefore, all powers of $x$ in the expansion are even.

1. Find the value of $r$ for the term independent of $x$

The general term is $T_{r+1} = \binom{6}{r}(x^4)^{6-r}\left(-\frac{1}{kx^2}\right)^r = \binom{6}{r}(-1)^r k^{-r} x^{24-6r}$.
For the term independent of $x$, the power of $x$ must be $0$:
$24 - 6r = 0$
$6r = 24$
$r = 4$

2. Equate the term to $\frac{5}{27}$ and solve for $k$

Substitute $r = 4$ into the coefficient part of the general term:
$T_5 = \binom{6}{4}(-1)^4 k^{-4} = 15k^{-4}$
Given that this term is $\frac{5}{27}$:
$15k^{-4} = \frac{5}{27}$
$k^{-4} = \frac{5}{27 \times 15}$
$k^{-4} = \frac{1}{81}$
$k^4 = 81$
Since $k$ is a positive constant, $k = 3$.

1. Identify the terms needed from the expansion of $\left(x^{4}-\frac{1}{3 x^{2}}\right)^{6}$

To find the term independent of $x$ in $\left(2-3x^{6}\right)\left(x^{4}-\frac{1}{3 x^{2}}\right)^{6}$, we need to multiply $2$ by the term independent of $x$ in the expansion, and $-3x^{6}$ by the term in $x^{-6}$ in the expansion.

2. Find the term in $x^{-6}$ in the expansion of $\left(x^{4}-\frac{1}{3 x^{2}}\right)^{6}$

The general term is $T_{r+1} = \binom{6}{r}(x^{4})^{6-r}\left(-\frac{1}{3x^{2}}\right)^{r} = \binom{6}{r}\left(-\frac{1}{3}\right)^{r}x^{24-6r}$.
For the term in $x^{-6}$, we set the power of $x$ to $-6$:
$24 - 6r = -6$
$6r = 30 \implies r = 5$
The term in $x^{-6}$ is $\binom{6}{5}\left(-\frac{1}{3}\right)^{5}x^{-6} = 6\left(-\frac{1}{243}\right)x^{-6} = -\frac{2}{81}x^{-6}$.

3. Calculate the final term independent of $x$

From part (b), the term independent of $x$ is $\frac{5}{27}$.
The term independent of $x$ in the product is:
$2\left(\frac{5}{27}\right) - 3\left(-\frac{2}{81}\right)$
$= \frac{10}{27} + \frac{6}{81}$
$= \frac{10}{27} + \frac{2}{27}$
$= \frac{12}{27} = \frac{4}{9}$