How do you simplify #root(3)(27b^18c^12)#?

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Answer 1 · 2018-04-08T19:00:32

#3b^6c^4#

27 is a perfect cube. #3^3=27#, so #root3(27)=3#

#root3(3*3*3*b^6*b^6*b^6*c^4*c^4*c^4)#

#root3(b^(6+6+6))=root3(b^6*b^6*b^6) rarr# Using rules of exponents, you can separate the variables and their exponents.

Perfect cubes are the results of a number being multiplied by itself twice.

#b^6# and #c^4# are being multiplied by themselves two times.

Take them out of the cube root:

#3b^6c^4#