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

2 Answers
Apr 24, 2018

the simplified answer would be -4

Explanation:

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

Apr 24, 2018

#-4#

Explanation:

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.