How do you simplify #\root [ 3] { 32x ^ { 12} y z ^ { 6} }#?

1 Answer
Jacob T. · Anjali G
May 31, 2017

#2 root [3] (4) * x^4 * root[3] (y) * z^2#

Explanation:

#root[3] (32x^12yz^6)#
#=(2^5*x^12*y*z^6)^(1/3)#
#=2^(5/3)*x^(12/3)*y^(1/3)*z^(6/3)#
#=2*2^(2/3)*x^4*y^(1/3)*z^2#

#=2 root [3] (4) * x^4 * root[3] (y) * z^2#

# = 2^(5/3)x^4y^(1/3)z^2#

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