We are given that $a$ and $b$ are whole numbers such that $a^b = 121$. We want to find the value of $(a-1)^{b+1}$.

1. Problem Description

We are given that $a$ and $b$ are whole numbers such that $a^b = 121$. We want to find the value of $(a-1)^{b+1}$.

2. Solution Steps

Since $a$ and $b$ are whole numbers and $a^b = 121$, we need to find the pairs of whole numbers $(a, b)$ that satisfy this equation. We know that $121 = 11^2$. We can also write $121 = 121^1$.

Case 1: $a = 11$ and $b = 2$.

Then $(a-1)^{b+1} = (11-1)^{2+1} = (10)^3 = 1000$.

Case 2: $a = 121$ and $b = 1$.

Then $(a-1)^{b+1} = (121-1)^{1+1} = (120)^2 = 14400$.

Since the problem does not provide enough information to determine which case is the correct one, we can assume that the intended solution corresponds to the smaller base $a$. Therefore we assume that $a=11$ and $b=2$.

$(a-1)^{b+1} = (11-1)^{2+1} = (10)^{3} = 1000$

3. Final Answer

1000