If $a = 3$ and $b =-2$, find the values of:$a^a+ b^b$

Given:

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

To do:

We have to find the value of $a^a+ b^b$.

Solution:

We know that,

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

Therefore,

$a^a+ b^b=(3)^3+(-2)^{-2}$

$=27+\frac{1}{(-2^{2})}$

$=27+\frac{1}{4}$

$=\frac{27\times4+1}{4}$

$=\frac{108+1}{4}$

$=\frac{109}{4}$

Hence, $a^a+b^b=\frac{109}{4}$.