Answer:
![a^(-51) = \frac1{a^(51)}](https://img.qammunity.org/2023/formulas/mathematics/high-school/zmmdkpzwj0tr5pllr82omzggbbme4akg2u.png)
Explanation:
When you're multiplying two powers with the same base (in our case, a) you add together the exponents.
![((a^1^4\cdot a^-^6)/(a^(25)))^3 =\\(\frac {a^(14+(-5))}{a^2^5})^3= (\frac {a^(14-6)}{a^2^5})^3 = (\frac {a^(8)}{a^2^5})^3](https://img.qammunity.org/2023/formulas/mathematics/high-school/mm4ol23au1cslhtnazrova798z9ol2kfvn.png)
Now, when you are dividing two powers with the same base (in our case, a again) you subtract the exponents:
![= (a^(8-25))^3 = (a^(-17))^3](https://img.qammunity.org/2023/formulas/mathematics/high-school/mz5cdynwvxlj1isowd91ljrmyjbn4f8jr4.png)
Finally, when you're calculating the power of a power, you multiply the exponents together:
![a^(-17\cdot 3) = a^(-51) = \frac1{a^(51)}](https://img.qammunity.org/2023/formulas/mathematics/high-school/ev230ksgjjyjockkmm2det6zmwtskzi465.png)
At this point you just have to choose - or check with your book/teacher if you prefer a negative exponent, or a fraction.