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Question

$6^{- 3} \times 6^{3} =$
$6^{0}$
1
none of these
both a & b above

A

1

B

both a & b above

C

6^{0}

D

none of these

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Solution

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$6^{- 3} \times 6^{3} = \frac{1}{6^{3}} \times 6^{3} = 1 = 6^{0}$ (Using $a^{- b} = \frac{1}{a^{b}} & a^{0} = 1$ )
Answer:- $C$

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