Question

Question 3(iv)
Find the value of:

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

Answer 1

$(3^{-1}+4^{-1}+5^{-1}{)}^0={(\frac{1}{3}+\frac{1}{4}+\frac{1}{5})}^0[\because a^{-m}=\frac{1}{a^m}]{\left(\frac{20+15+12}{60}\right)}^0={\left(\frac{47}{60}\right)}^0=1[\because a^0=1]$

Alternative:

$(3^{-1}+4^{-1}+5^{-1}{)}^0$
= 1 $[\because a^0=1]$ (Anything to the power zero is equal to 1.)