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

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Answer 1 · 2017-10-09T00:38:37

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

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

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

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

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