Question #45348

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Answer 1 · 2017-03-17T16:11:29

#16^(2/3)# can be rewritten as #root(3) (16^2)#

Solve for the inside first:

#16^2=256#

Now take the third root of this. Break down 256 into powers of 2:

#256=2^8#

#8# goes into #3# 2 times and you have a remainder of #2^2=4# So we have:

#2*2 root(3)(4)#

Answer 2 · 2017-03-17T16:30:35

#4root(3)(2^2)#

Before you do anything else it is wise to look and see if you can spot any short cuts. I have spotted 1

Just accept that #16^(2/3)# is another way of writing #root(3)(16^2)#

When ever you have a root check to see if there are any values you can 'extract' from it. In this case we would be looking for cubed value.

#16=4^2=2^2xx2^2 = 2^3xx2# so

#16^2=2^3xx2xx2^3xx2=2^3xx2^3xx2^2#

#root(3)(16^2)->root(3)(2^3xx2^3xx2^2)" "=" "2xx2xxroot(3)(2^2)" "=" "4root(3)(2^2)#