How do you simplify #8^(2/3) * 8#?

1 Answer
Oct 30, 2015

Use your knowledge of indices

Ans: #8^(5/3)# or #root(3)(8^5)# or #32#

Explanation:

#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#