1. **State the problem:** We have a geometric sequence with common ratio $r$. The difference between the 2nd term $a r$ and the 5th term $a r^4$ is 156. The difference between the 7th term $a r^6$ and the 4th term $a r^3$ is 1404. We need to find all possible values of the common ratio $r$. 2. **Set up the equations:** From the problem, $$a r^4 - a r = 156$$ $$a r^6 - a r^3 = 1404$$ 3. **Factor out $a r$ in both:** $$a r (r^3 - 1) = 156$$ $$a r^3 (r^3 - 1) = 1404$$ 4. **Divide the second equation by the first:** $$\frac{a r^3 (r^3 - 1)}{a r (r^3 - 1)} = \frac{1404}{156}$$ Simplify numerator and denominator: $$\frac{r^3}{r} = 9$$ $$r^{2} = 9$$ 5. **Solve for $r$:** $$r = \pm 3$$ 6. **Find $a$ for each $r$:** Using the first equation: $$a r (r^3 - 1) = 156$$ If $r=3$: $$a \times 3 (27 - 1) = 156 \implies a \times 3 \times 26 = 156 \implies 78a = 156 \implies a = 2$$ If $r=-3$: $$a \times (-3) ((-3)^3 -1) = 156 \implies a \times (-3)(-27 -1) = 156 \implies a \times (-3)(-28) = 156 \implies 84a = 156 \implies a = \frac{156}{84} = \frac{13}{7}$$ 7. **Final answer:** The possible values for the common ratio are: $$r = 3 \quad \text{and} \quad r = -3$$