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Question
Simplify:
`(8^(1/3) xx 16^(1/3))/(32^(-1/3))`
Solution
`(8^(1/3) xx 16^(1/3))/(32^(- 1/3)) = ((2^3)^(1/3) xx (2^4)^(1/3))/((2^5)^(-1/3))` ...[∵ (am)n = amn]
= `(2^(3 xx 1/3) xx 2^(4 xx 1/3))/(2^(5 xx -1/3))`
= `(2^(3/3 + 4/3))/(2^(-5/3))` ...`[∵ a^m/a^n = a^(m - n)]`
= `2^(7/3)/(2^(-5/3))`
= `2^(7/3 + 5/3)`
= `2^(12/3)`
= 24
= 16
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