How do you simplify #2\root[ 3] { 16x ^ { 2} y ^ { 7} }#?

1 Answer
sankarankalyanam · Manikandan S.
Oct 9, 2017

#=root(3)(2^7x^2y^7)=2^(7/3)x^(2/3)y^(7/3)#

Explanation:

#2*(16x^2y^7)^(1/3)=2*16^(1/3)*x^(2/3)*y^(7/3)#
#=2*2^(4/3)*x^(2/3)*y^(7/3)#
#=2^(7/3)*x^(2/3)*y^(7/3)#
#=root(3)(2^7x^2y^7))#

Aliter :
#=root(3)2^(3)*root(3)(2^4*x^2*y^7)#
#=root(3)(2^3*2^4*x^2*y^7)#
#=root(3)(2^7x^2y^7)#

#=4y^2root(3)(2x^2y)#

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