#8^(2/3)# can be written as #root(3)(8^2)#.
If you know that #8^2# is #64#, and that cube root of #64# (#root(3)64)# is #4#, then this it is easy to solve the problem in this way.
You could also write #8^(2/3)# as #root(3)(8*8)#.
Take the cube root of each #8# separately and multiply them
#8^(2/3)*8#
(#root(3)8*root(3)8#)*8
#(2*2)*8#
#4*8#
#32#
Another way to solve this:
When two numbers are being multiplied (#8^(2/3)# and #8#) and their bases (#8#) are the same, the product can simply be written as the base (#8#) to the power of the addition of the powers (#2/3+1#)
#8^(2/3)*8^1#
#8^(2/3+1)#
#8^(5/3)#
#8^(5/3)# can be written as #root(3)(8^5)#
#root(3)(8^5)#
#root(3)(8*8*8*8*8)#
#2*2*2*2*2#
#32#
