How do you write the simplified form of #-64^(1/ 3)#?

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Answer 1 · 2018-04-24T04:24:43

the simplified answer would be -4

Let's factor out 64:
#64=2^6#
#-(2^6)^(1/3)#
#=-2^(6.(1/3))#
#=-2^2#
#=-4#

Answer 2 · 2018-04-24T06:35:58

#-4#

Recall one of the laws of indices:

#sqrtx = x^(1/2)" "and " "root3(x) = x^(1/3)#

#-64^(1/3) = root3(-64)#

#64# is a perfect cube: #64=4^3#

#root3(-64) =-4#

You could also work with the prime factors:

#root3(-64) = root3(-(2^6))#

#=-2^2#

#=-4#

Note that perfect cubes can be negative, but perfect squares cannot.