If the coefficient of $$x^{15}$$ in the expansion of $$\left(ax^3 + \frac{1}{bx^{\frac{1}{3}}}\right)^{15}$$ is equal to the coefficient of $$x^{-15}$$ in the expansion of $$\left(ax^{\frac{1}{3}} - \frac{1}{bx^3}\right)^{15}$$, where $$a$$ and $$b$$ are positive real numbers, then for each such ordered pair $$a, b$$:

Step 1: Find the coefficient of $$x^{15}$$ in the first expansion

The first binomial expression is:

$$\left(ax^3 + \frac{1}{bx^{\frac{1}{3}}}\right)^{15}$$

The general term $$T_{r+1}$$ in this expansion is given by:

$$T_{r+1} = \binom{15}{r} (ax^3)^{15-r} \left(\frac{1}{bx^{\frac{1}{3}}}\right)^r$$

$$T_{r+1} = \binom{15}{r} a^{15-r} b^{-r} x^{3(15-r) - \frac{r}{3}}$$

$$T_{r+1} = \binom{15}{r} a^{15-r} b^{-r} x^{45 - 3r - \frac{r}{3}}$$

$$T_{r+1} = \binom{15}{r} a^{15-r} b^{-r} x^{45 - \frac{10r}{3}}$$

To find the coefficient of $$x^{15}$$, we set the exponent of $$x$$ equal to 15:

$$45 - \frac{10r}{3} = 15$$

$$\frac{10r}{3} = 30$$

$$10r = 90$$

$$r = 9$$

Substituting $$r = 9$$ back into the term gives the coefficient of $$x^{15}$$:

$$\text{Coefficient of } x^{15} = \binom{15}{9} a^{15-9} b^{-9} = \binom{15}{9} \frac{a^6}{b^9}$$

Step 2: Find the coefficient of $$x^{-15}$$ in the second expansion

The second binomial expression is:

$$\left(ax^{\frac{1}{3}} - \frac{1}{bx^3}\right)^{15}$$

The general term $$T_{k+1}$$ in this expansion is given by:

$$T_{k+1} = \binom{15}{k} \left(ax^{\frac{1}{3}}\right)^{15-k} \left(-\frac{1}{bx^3}\right)^k$$

$$T_{k+1} = \binom{15}{k} a^{15-k} (-1)^k b^{-k} x^{\frac{15-k}{3} - 3k}$$

$$T_{k+1} = \binom{15}{k} a^{15-k} (-1)^k b^{-k} x^{5 - \frac{k}{3} - 3k}$$

$$T_{k+1} = \binom{15}{k} a^{15-k} (-1)^k b^{-k} x^{5 - \frac{10k}{3}}$$

To find the coefficient of $$x^{-15}$$, we set the exponent of $$x$$ equal to -15:

$$5 - \frac{10k}{3} = -15$$

$$20 = \frac{10k}{3}$$

$$60 = 10k$$

$$k = 6$$

Substituting $$k = 6$$ back into the term gives the coefficient of $$x^{-15}$$:

$$\text{Coefficient of } x^{-15} = \binom{15}{6} a^{15-6} (-1)^6 b^{-6} = \binom{15}{6} \frac{a^9}{b^6}$$

Step 3: Equate the two coefficients and solve for the relationship

We are given that the two coefficients are equal:

$$\binom{15}{9} \frac{a^6}{b^9} = \binom{15}{6} \frac{a^9}{b^6}$$

Using the identity $$\binom{n}{r} = \binom{n}{n-r}$$, we know that $$\binom{15}{9} = \binom{15}{15-9} = \binom{15}{6}$$. Therefore, these terms cancel out from both sides:

$$\frac{a^6}{b^9} = \frac{a^9}{b^6}$$

Cross-multiplying the terms gives:

$$a^6 b^6 = a^9 b^9$$

$$1 = a^3 b^3$$

$$(ab)^3 = 1$$

Since $$a$$ and $$b$$ are positive real numbers, taking the cube root of both sides gives:

$$ab = 1$$

Conclusion:

the correct option is $$ab = 1$$.