How do you simplify #root4(64u^7v^7)#?

1 Answer
mason m
Nov 30, 2015

#2uvroot4(4u^3v^3)#

Explanation:

Know that: #{(root(4)(a^4)=a),(root(4)(ab)=root(4)a*root4b):}#

We can rewrite #root4(64u^7v^7)# as #root4(2^4*2^2*u^4*u^3*v^4*v^3)#—from which we can simplify to #root4(2^4)*root4(2^2)*root4(u^4)*root4(u^3)*root4(v^4)*root4(u^4)*root4(v^3)#.

From there, we can move all the terms raised to the #4"th"# power using the first rule.

#2uv*root4(2^2)*root4(u^3)*root4(v^3)#

Work in reverse and recombine terms:

#2uvroot4(4u^3v^3)#

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