If $a = 3$ and $b =-2$, find the values of:$(a+b)^{ab}$

Given:

$a = 3$ and $b =-2$

To do:

We have to find the value of $(a+ b)^{ab}$.

Solution:

We know that,

$a^{-m}=\frac{1}{a^m}$

Therefore,

$(a+ b)^{ab}=[3+(-2)]^{3\times(-2)}$

$=(3-2)^{-6}$

$=(1)^{-6}$

$=\frac{1}{1^6}$

$=\frac{1}{1}$

$=1$

Hence, $(a+b)^{ab}=1$.