Help with simple algebra equation I am doing some math to prepare for a test, and $$\left(\frac{a^2}{b^{-1}}\right)^{-3}\cdot\left(\frac{(2b)^3}{a^{-3}}\right)$$ is apparently $8a^{-3}$. When I try to solve it the closest I get is $7a^{-3}$. After multiplying the upper case numbers and dissolving the parentheses I am left with: $$\frac{a^{-6}\cdot6b^{3}}{b^3\cdot a^{-3}}$$ I get that when I cross out the $b^3$'s I am left with $1$ since $b^0 = 1$, but that is just $6+1$. With the a's left I get $7a^{-3}$. Could anyone tell me what I'm missing, how does it turn to $8a^{-3}$?      

Actually the first mistake is that $$(2b)^3=2^3b^3=8b^3$$ and not $6b^3$ as you have it. Moreover, when you simplify the $b^3$ you have to multiply with $1$ and not add $1$. This will give you the required result.
