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Question
Simplify:
`(9^(1/3) xx 27^(-1/2))/(3^(1/6) xx 3^(- 2/3))`
Solution
`(9^(1/3) xx 27^(-1/2))/(3^(1/6) xx 3^(-2/3)) = ((3^2)^(1/3) xx (3^3)^(-1/2))/(3^(1/6) xx 3^(-2/3))` ...[∵ (am)n = amn]
= `(3^(2/3) xx 3^(-3/2))/(3^(1/6) xx 3^(-2/3))` ...[∵ am × an = am + n]
= `(3^(2/3 - 3/2))/(3^(1/6 - 2/3))`
= `(3^((4 - 9)/6))/(3^((1 - 4)/6))` ...`[∵ a^m/a^n = a^(m - n)]`
= `(3^(-5/6))/(3^(-3/6)`
= `3^(- 5/6 + 3/6)`
= `3^(-2/6)`
= `3^(-1/3)`
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