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Question

Question 3(iv)
Find the value of:

$\left(3^{-1}+4^{-1}+5^{-1}\right)^0$

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Solution

$\left(3^{-1}+4^{-1}+5^{-1}\right)^0=\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)^0\:\:\:\:\left[\because a^{-m}=\frac{1}{a^m}\right]$


$ =\left(\frac{20+15+12}{60}\right)^0$


$=\left(\frac{47}{60}\right)^0$


$=1\:\:\:\left[\because a^0=1\right]$

(Anything to the power zero is equal to 1)


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