Explain why the binomial expansion of $\left(ax + \dfrac{2}{x}\right)^6$ contains only even powers of $x$. Write down, and simplify, the first three terms in the expansion of $\left(1 - \dfrac{x}{a}\right)^3$ in ascending powers of $x$. Given that the coefficient of $\dfrac{1}{x^4}$ in the expansion of $\left(ax + \dfrac{2}{x}\right)^6$ is 1728 times the coefficient of $x^2$ in the expansion of $\left(1 - \dfrac{x}{a}\right)^3$, find the value of $a$. Using the value of $a$ found in (c), show that there is no term independent of $x$ for $\left(1 - \dfrac{x}{a}\right)^3 \left(1 + \dfrac{1}{x}\right)$.

1. Write down the general term of the binomial expansion

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

2. Simplify the expression to find the power of $x$ and explain why it is always even

$T_{r+1} = \binom{6}{r} a^{6-r} 2^r x^{6-r} x^{-r}$
$T_{r+1} = \binom{6}{r} a^{6-r} 2^r x^{6-2r}$
The power of $x$ is $6 - 2r$. Since $r$ is an integer ($0 \leqslant r \leqslant 6$), $2r$ is an even integer. The difference between two even integers ($6$ and $2r$) is always an even integer. Therefore, the expansion contains only even powers of $x$.

1. Apply the binomial expansion formula for the first three terms

$\left(1 - \dfrac{x}{a}\right)^3 = 1^3 + \binom{3}{1}(1)^2\left(-\dfrac{x}{a}\right) + \binom{3}{2}(1)^1\left(-\dfrac{x}{a}\right)^2 + \dots$

2. Simplify the terms

$= 1 + 3\left(-\dfrac{x}{a}\right) + 3\left(\dfrac{x^2}{a^2}\right) = 1 - \dfrac{3}{a}x + \dfrac{3}{a^2}x^2$

1. Find the value of $r$ for the term in $\frac{1}{x^4}$ in the expansion of $\left(ax + \frac{2}{x}\right)^6$

The general term is $T_{r+1} = \binom{6}{r} (ax)^{6-r} \left(\frac{2}{x}\right)^r = \binom{6}{r} a^{6-r} 2^r x^{6-2r}$.
For the term in $\frac{1}{x^4}$ (or $x^{-4}$):
$6 - 2r = -4$
$2r = 10 \implies r = 5$

2. Calculate the coefficient of $\frac{1}{x^4}$

Substitute $r = 5$ into the coefficient part:
Coefficient $= \binom{6}{5} a^{6-5} 2^5 = 6 \cdot a \cdot 32 = 192a$

3. Identify the coefficient of $x^2$ in the expansion of $\left(1 - \frac{x}{a}\right)^3$

From the expansion of $\left(1 - \frac{x}{a}\right)^3 = 1 - 3\left(\frac{x}{a}\right) + 3\left(\frac{x}{a}\right)^2 - \dots$,
the coefficient of $x^2$ is $\frac{3}{a^2}$.

4. Form an equation using the given relationship and solve for $a$

$192a = 1728 \left(\frac{3}{a^2}\right)$
$192a^3 = 5184$
$a^3 = 27$
$a = 3$

1. Find the value of $a$ from part (c) to use in the expression.

From part (c), the coefficient of $\frac{1}{x^4}$ in $\left(ax + \frac{2}{x}\right)^6$ is $\binom{6}{5}(ax)^1\left(\frac{2}{x}\right)^5 = 6a(32) = 192a$.
The coefficient of $x^2$ in $\left(1 - \frac{x}{a}\right)^3$ is $\binom{3}{2}(1)^1\left(-\frac{x}{a}\right)^2 = \frac{3}{a^2}$.
Given $192a = 1728 \left(\frac{3}{a^2}\right) \implies 192a^3 = 5184 \implies a^3 = 27 \implies a = 3$.

2. Substitute $a=3$ into the expression and expand the first bracket.

Substitute $a = 3$ into $\left(1 - \frac{x}{a}\right)^3 \left(1 + \frac{1}{x}\right)$ to get $\left(1 - \frac{x}{3}\right)^3 \left(1 + \frac{1}{x}\right)$.
Expanding $\left(1 - \frac{x}{3}\right)^3 = 1 - 3\left(\frac{x}{3}\right) + 3\left(\frac{x}{3}\right)^2 - \left(\frac{x}{3}\right)^3 = 1 - x + \frac{x^2}{3} - \frac{x^3}{27}$.

3. Multiply the expansions to find the term independent of $x$.

We need the term independent of $x$ in $\left(1 - x + \frac{x^2}{3} - \frac{x^3}{27}\right) \left(1 + \frac{1}{x}\right)$.
The term independent of $x$ is formed by multiplying the constant term in the first bracket by the constant term in the second bracket, and the $x$ term in the first bracket by the $\frac{1}{x}$ term in the second bracket.
Term independent of $x$ = $(1)(1) + (-x)\left(\frac{1}{x}\right) = 1 - 1 = 0$.
Since the coefficient is 0, there is no term independent of $x$.