simplify: 9(a^-2)b over (b^-1)((8a)^-1/3))

Question

anonymous · Answer

can't really make sense of this

darthsid · Answer

$$\frac{9a^{-2}b}{b^{-1}(8a)^{-\frac{1}{3}}}$$
is this it?

anonymous · Answer

yeah its this:
$$(9a ^{-2}b) \div (b ^{-1}\sqrt[3]{8a})$$

anonymous · Answer

when i tried to simplify it i got $$9b ^{2} \div (2a ^{7/3})$$

anonymous · Answer

$$\frac{9 (8a)^{\frac{1}{3}} a^{(-2)}b^2}{1}$$

anonymous · Answer

$$\frac{9 *2 (a)^{\frac{1}{3}} a^{(-2)}b^2}{1}$$
$$\frac{18 (a)^{-\frac{5}{3}}b^2}{1}$$

anonymous · Answer

There are two different problems here.  imranmeah has a negative 1/3 on the bottom which brings it to the top...but sunyah has a cube root on the bottom (which would stay there). 

Is it $$\frac{9a^{-2}b}{b^{-1}(8a)^{1/3}}$$or$$\frac{9a^{-2}b}{b^{-1}(8a)^{-1/3}}$$?

anonymous · Answer

it is just a cube root- that can be changed to (1/3) but not NEGATIVE... oohhh i see my mistake. :) THANK YOU!

anonymous · Answer

Then your answer of $$\frac{9b^{2}}{2a^{7/3}}$$ would be right.

anonymous · Answer

thank youu i must have typed it wrong into wolfram alpha to check, got imranmeah91's answer, and was super confused because my work on paper seemed right. :)

anonymous · Answer

that's why i always trust my own pencil :)