How do you simplify #root4(32x^4 y^5 n^10)#?

1 Answer
Jul 9, 2016

Answer:

#root(4)(32x^4y^5n^10)=2xyn^2root(4)(2yn^2)#

Explanation:

#root(4)(32x^4y^5n^10)#

= #root(4)(2×2×2×2×2×x×x×x×x×y×y×y×y×y×n×n×n×n×n×n×n×n×n×n)#

= #root(4)(ul(2×2×2×2)×2×ul(x×x×x×x)×ul(y×y×y×y)×y×ul(n×n×n×n)×ul(n×n×n×n)×n×n)#

= #2x×y×n×n×root(4)(2y×n×n)#

= #2xyn^2root(4)(2yn^2)#