Subtract two exponents with same base How do I solve

First I thought the difference is $D - A = 9^{12} - 9^9$, but I didn't know the rules for subtracting exponents with different bases.
But the question states that the numbers are evenly spaced, so I guess I need to calculate $9^{13} - 9^{11}$. Since I know that the distance between A and D is 1.5 times the distance between C and E, then the correct answer must be $\frac{3}{2} \cdot (9^{13} - 9^{11})$.
Would it be easier if I say something like
$$
\frac{3}{2} \cdot (9^{13} - 9^{11}) = \frac{3}{2} \cdot (9^3 \cdot 9^{10} - 9 \cdot 9^{10}) = \frac{3}{2} \cdot (9^3 - 9) \cdot 9^{10} = \frac{3}{2} \cdot 720 \cdot 9^{10}
$$
Edit
I guess what I should have done was to say
$$
\frac{3}{2} \cdot (9^{13} - 9^{11}) = \frac{3}{2} \cdot (9^{2} \cdot 9^{11} - 9^{11}) = \frac{3}{2} \cdot (9^{2} - 1) \cdot 9^{11} = \frac{3}{2} \cdot 80 \cdot 9^{11} = 120 \cdot 9^{11}
$$
which is the fifth (last) answer in the image.

Yes, it seems you have the correct answer, but it can be done even more simply:
$$\frac32(9^{13}-9^{11}) = \frac32(9^2\cdot9^{11} - 1\cdot9^{11}) = \frac32\cdot 9^{11}\cdot (81-1)=80\cdot\frac32\cdot9^{11} = 120\cdot 9^{11}$$
