Solve: Simplifying 14a^-3 ÷ 7a^-3 with Like Terms

Due to the fact that the numerator and the denominator of the fraction have terms with identical bases, we will begin by applying the law of exponents for the division of terms with identical bases:

$\frac{b^m}{b^n}=b^{m-n}$We apply it to the problem:

$\frac{14a^{-3}}{7a^{-3}}=2a^{-3-(-3)}=2a^{-3+3}=2a^0$In the first step we simplify the numerical part of the fraction. This is a simple and intuitive step as it makes it easier to work with the said fraction.:

$\frac{14a^{-3}}{7a^{-3}}=\frac{14}{7}\cdot\frac{a^{-3}}{a^{-3}}=2a^{-3-(-3)}=\ldots$We then return to the problem and remember that any number raised to the 0th power is 1, that is:

$b^0=1$Thus, in the problem we obtain the following:

$2a^0=2\cdot1=2$Therefore, the correct answer is option B.