What is $\left(\frac{7}{8}\right)^3 \cdot \left(\frac{7}{8}\right)^{-3}$?
By definition, if $a$ is nonzero, then $a^{-3}$ is the reciprocal of $a^3$.  So, $\left(\frac78\right)^3$ and $\left(\frac78\right)^{-3}$ are reciprocals.  Therefore, their product is $\boxed{1}$.
1
Level 3
Prealgebra
Problem: What is $\left(\frac{7}{8}\right)^3 \cdot \left(\frac{7}{8}\right)^{-3}$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the expression, we first recognize the properties of exponents. The expression involves multiplying two terms with the same base, which is $\frac{7}{8}$. According to the laws of exponents, when multiplying terms with the same base, we add the exponents. 

The first term is $\left(\frac{7}{8}\right)^3$, which has an exponent of $3$. The second term is $\left(\frac{7}{8}\right)^{-3}$, which has an exponent of $-3$. 

Now, we can combine the exponents:
$$
3 + (-3) = 3 - 3 = 0.
$$

Thus, we can rewrite the expression as:
$$
\left(\frac{7}{8}\right)^{3 + (-3)} = \left(\frac{7}{8}\right)^0.
$$

Next, we apply the property of exponents that states any non-zero number raised to the power of $0$ equals $1$. Since $\frac{7}{8}$ is a non-zero number, we conclude:
$$
\left(\frac{7}{8}\right)^0 = 1.
$$

Therefore, the final answer is $1$.<|END_OF_LATENT|> The final answer is $1$.