Simplify #root(3)(48b^10)#
Find factors of #48# that are perfect cubes
(the cube of 2 is #2^3=8#, the cube of 3 is #3^3=27#, etc).
#48 = 8 *6 = 2^3 *6#
The cube root of #2^3# is #2#. The#2^3# comes "out" of the cube root symbol as a #2#, while the #6# stays inside.
#root(3)(48b^10)=root(3)(8*6b^10)=root(3)(color(red)(2^3)*6b^10)=color(red)2root(3)(6b^10)#
Next lets look at the cube root of #b^10#.
If you divide #10# by #3#, the result is #3# with a remainder of #1#.
In other words #b^10=b^3*b^3*b^3*b^1#. Each #b^3# is a perfect cube, and comes "out" of the cube root symbol as #b#.
#2root(3)(6b^10)=2root(3)(6color(red)(b^3)*color(blue)(b^3)*color(magenta)(b^3)*b^1)=2color(red)b*color(blue)b*color(magenta)broot(3)(6b)#
Multiplying the three #b#'s outside the cube root symbol gives
#2b^3root(3)(6b)#
