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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, $(a+b)^{ab}=1$.
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