Find the first 4 terms in the expansion of $\left(\frac{a}{x}+3x^2\right)^6$ in ascending power of $x$, simplifying each term. Given that there is no term in $x^3$ in the expansion of $(2x^3-1)\left(\frac{a}{x}+3x^2\right)^6$, find the value of the positive constant $a$.

1. Apply the binomial theorem to expand the first 4 terms

$\left(\frac{a}{x}+3x^2\right)^6 = \binom{6}{0}\left(\frac{a}{x}\right)^6(3x^2)^0 + \binom{6}{1}\left(\frac{a}{x}\right)^5(3x^2)^1 + \binom{6}{2}\left(\frac{a}{x}\right)^4(3x^2)^2 + \binom{6}{3}\left(\frac{a}{x}\right)^3(3x^2)^3 + \dots$

2. Simplify the first two terms

Term 1: $1 \cdot \frac{a^6}{x^6} \cdot 1 = \frac{a^6}{x^6}$
Term 2: $6 \cdot \frac{a^5}{x^5} \cdot 3x^2 = \frac{18a^5}{x^3}$

3. Simplify the third and fourth terms and state the final expansion

Term 3: $15 \cdot \frac{a^4}{x^4} \cdot 9x^4 = 135a^4$
Term 4: $20 \cdot \frac{a^3}{x^3} \cdot 27x^6 = 540a^3 x^3$

The first 4 terms are $\frac{a^6}{x^6} + \frac{18a^5}{x^3} + 135a^4 + 540a^3 x^3$.

1. Identify the terms from the expansion of $\left(\frac{a}{x}+3x^2\right)^6$ needed to form the $x^3$ term in the product.

To find the $x^3$ term in $(2x^3-1)\left(\frac{a}{x}+3x^2\right)^6$, we need the constant term ($x^0$) and the $x^3$ term from the expansion of $\left(\frac{a}{x}+3x^2\right)^6$.

The general term is $T_{r+1} = \binom{6}{r} \left(\frac{a}{x}\right)^{6-r} (3x^2)^r = \binom{6}{r} a^{6-r} 3^r x^{3r-6}$.

For the constant term ($x^0$):
$3r - 6 = 0 \implies r = 2$
Term $= \binom{6}{2} a^4 (3)^2 = 15 \times 9 a^4 = 135a^4$

For the $x^3$ term:
$3r - 6 = 3 \implies r = 3$
Term $= \binom{6}{3} a^3 (3)^3 x^3 = 20 \times 27 a^3 x^3 = 540a^3 x^3$

2. Form an equation for the coefficient of $x^3$ and solve for $a$.

The $x^3$ term in the product is:
$(2x^3)(135a^4) + (-1)(540a^3 x^3) = (270a^4 - 540a^3)x^3$

Since there is no term in $x^3$, the coefficient is zero:
$270a^4 - 540a^3 = 0$
$270a^3(a - 2) = 0$

Since $a$ is a positive constant, $a \neq 0$, so:
$a = 2$