Let $$a, b$$ be two non-zero real numbers. If $$p$$ and $$r$$ are the roots of the equation $$x^2 - 8ax + 2a = 0$$ and $$q$$ and $$s$$ are the roots of the equation $$x^2 + 12bx + 6b = 0$$, such that $$\dfrac{1}{p}, \dfrac{1}{q}, \dfrac{1}{r}, \dfrac{1}{s}$$ are in A.P., then $$a^{-1} - b^{-1}$$ is equal to ______.

Given equations:

$$x^2-8ax+2a=0$$

with roots $$p,r$$

and

$$x^2+12bx+6b=0$$

with roots $$q,s$$

Using Vieta’s formulas,

$$p+r=8a,\qquad pr=2a$$

Therefore,

$$\frac1p+\frac1r=\frac{p+r}{pr}=\frac{8a}{2a}=4$$

Similarly,

$$q+s=-12b,\qquad qs=6b$$

Hence,

$$\frac1q+\frac1s=\frac{q+s}{qs}=\frac{-12b}{6b}=-2$$

Now let

$$\frac1p,\ \frac1q,\ \frac1r,\ \frac1s$$

be four consecutive terms of an A.P.

Assume them as

$$X_1,\ X_1+d,\ X_1+2d,\ X_1+3d$$

Then,

$$X_1+X_3=4$$

$$X_1+(X_1+2d)=4$$

$$2X_1+2d=4$$

$$X_1+d=2$$

So,

$$X_2=2$$

Also,

$$X_2+X_4=-2$$

$$2+(X_1+3d)=-2$$

Using

$$X_1+d=2,$$

$$2+(2+2d)=-2$$

$$4+2d=-2$$

$$d=-3$$

Hence,

$$X_1=2-(-3)=5$$

Therefore, the A.P. is

$$5,\ 2,\ -1,\ -4$$

Thus,

$$\frac1p=5,\qquad \frac1r=-1$$

and

$$\frac1q=2,\qquad \frac1s=-4$$

Now,

$$\frac1{pr}=\frac1p\cdot\frac1r=5(-1)=-5$$

Since

$$pr=2a,$$

$$\frac1{2a}=-5$$

$$a^{-1}=-10$$

Similarly,

$$\frac1{qs}=\frac1q\cdot\frac1s=2(-4)=-8$$

Since

$$qs=6b,$$

$$\frac1{6b}=-8$$

$$b^{-1}=-48$$

Therefore,

$$a^{-1}-b^{-1}=-10-(-48)=38$$

Hence, the required value is

$$\boxed{38}$$