Question 204657
 need to express [16(a^2*b^4)^(-1/2)] / [b-3] as a single fraction involving no negative powers.
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[16(a^2*b^4)^(-1/2)] = 1/[16(a^2*b^4)^(1/2)]
= 16/ab^2
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was b-3 supposed to be b^(-3) ?
If so:
1/b^(-3) = b^3
--> 16b/a