How do you simplify #sqrt(49x^12y^4z^8)#?

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Answer 1 · 2018-07-16T07:48:02

#7x^6y^2z^4#

#sqrt(49x^12y^4z^8)=sqrt((7x^6y^2z^4)^2)=7x^6y^2z^4#

Answer 2 · 2018-07-16T07:49:25

#sqrt(49x^12y^4z^8) = 7x^6y^2z^4#

Find the square root of each factor.

Divide the indices by #2#

#sqrt(49x^12y^4z^8) = 7x^6y^2z^4#

Only the principal (positive) root is required.