How do you solve #root3(3x)+6=10#?

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Answer 1 · 2018-06-14T14:03:48

#x=64/3#

We can take the original:

#root(3)(3x)+6=10#

and subtract 6 from both sides:

#root(3)(3x)=4#

#3x=64#

#x=64/3#

Answer 2 · 2018-06-14T14:04:36

#x=64/3#

#root3(3x)+6=10#

#root3(3x)=4#

#(root3(3x))^3=4^3#

#3x=64#

#x=64/3#