Question 65

$4^{0} + 5^{0} + 6^{0} = (4 + 5 + 6)^{0}$

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Solution

False
Here, $4^{0} + 5^{0} + 6^{0} = 1 + 1 + 1 = 3$
and $(4 + 5 + 6)^{0} = (15)^{0} = 1$
Hence, $4^{0} + 5^{0} + 6^{0} \neq (4 + 5 + 6)^{0}$

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