We are given that $a$ and $b$ are whole numbers such that $a^b = 121$. We need to evaluate $(a-1)^{b+1}$.

1. Problem Description

We are given that $a$ and $b$ are whole numbers such that $a^b = 121$. We need to evaluate $(a-1)^{b+1}$.

2. Solution Steps

First, we need to find possible values of $a$ and $b$ such that $a^b = 121$. Since $a$ and $b$ are whole numbers, we can write 121 as $11^2$ or $121^1$.

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

Then, we have to evaluate $(a-1)^{b+1} = (11-1)^{2+1} = (10)^3 = 1000$.

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

Then, we have to evaluate $(a-1)^{b+1} = (121-1)^{1+1} = (120)^2 = 14400$.

Since the problem does not state that $a$ and $b$ are non-zero, we can also consider the case $a=121$ and $b=1$.

$a=11$, $b=2$ are possible solutions.

$11^2 = 121$, so $a=11$ and $b=2$.

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

$121^1 = 121$, so $a=121$ and $b=1$.

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

If $a = 121$ and $b=1$, then $(a-1)^{b+1} = (121-1)^{1+1} = 120^2 = 14400$.

If $a = 11$ and $b=2$, then $(a-1)^{b+1} = (11-1)^{2+1} = 10^3 = 1000$.

3. Final Answer

The possible values for $(a-1)^{b+1}$ are 1000 and

1

4

4

0

0. Since the problem implies there is only one answer, we will proceed assuming the most obvious case $a=11$ and $b=2$.

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

Final Answer: The final answer is $\boxed{1000}$