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Question

Simplify: $[3^{8} \div 3^{6}] \times (4^{0} + 5^{0} + 6^{0})$

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Solution

$= \frac{3^{8}}{3^{6}} \times (1 + 1 + 1)$ since $a^{0} = 1$
$= 3^{8 - 6} \times 3$ since $\frac{a^{m}}{a^{n}} = a^{m - n}$
$= 3^{2} \times 3 = 9 \times 3 = 27$

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