How do you solve #root3x -4 = 2#?

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Answer 1 · 2016-05-02T11:00:55

#x=216#

Add 4 to both sides

#root(3)(x)+0=2+4#

#root(3)(x)=6#

Example: #(sqrt(4))^2 = 4# In squaring the square root I have reversed the process of square rooting 4

Using the above method

Cube both sides

#x=6^3#

#x=216#