hi. I'm expected to solve (27ab^3)^1/3 times the sq. root of (there is a three in the nook of the square root) 5a^4b. I follow it until it reduces to the square root of (still with the three in the nook), 3^3a^3b^3 times 5 a^2 b. Where does the three on the first a randomly come from? how does the a randomly get squared? Thanks for your help!!

Question

hi.  I'm expected to solve (27ab^3)^1/3 times the sq. root of (there is a three in the nook of the square root) 5a^4b.  I follow it until it reduces to  the square root of (still with the three in the nook), 3^3a^3b^3 times 5 a^2 b.  Where does the three on the first a randomly come from?  how does the a randomly get squared? Thanks for your help!!

Reiny · Answer

if your first factor had been (27(ab)^3)^1/3 that would become 3ab Did you forget a bracket?