How do you solve #2root3(10-3x)=root3(2-x)#?

First, cube each side of the equation to eliminate the radicals while keeping the equation balanced:

#(2root(3)(10 - 3x))^3 = (root(3)(2 - x))^3#

#2^3(10 - 3x) = 2 - x#

#8(10 - 3x) = 2 - x#

Next, eliminate the parenthesis on the left side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

#color(red)(8)(10 - 3x) = 2 - x#

#(color(red)(8) xx 10) - (color(red)(8) xx 3x) = 2 - x#

#80 - 24x = 2 - x#

Then, add #color(red)(24x)# and subtract #color(blue)(2)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#-color(blue)(2) + 80 - 24x + color(red)(24x) = -color(blue)(2) + 2 - x + color(red)(24x)#

#78 - 0 = 0 - 1x + color(red)(24x)#

#78 = (-1 + color(red)(24))x#

#78 = 23x#

Now, divide each side of the equation by #color(red)(23)# to solve for #x# while keeping the equation balanced:

#78/color(red)(23) = (23x)/color(red)(23)#

#78/23 = (color(red)(cancel(color(black)(23)))x)/cancel(color(red)(23))#

#78/23 = x#

#x = 78/23#

#2root3(10-3x)=root3(2-x)#

Cube both sides

#:.(2root3(10-3x))^3=(root3(2-x))^3#

#:.root3(a-b)*2root3(a-b)*root3(a-b)=a-b#

#:.2^3(10-3x)=2-x#

#:.8(10-3x)=2-x#

#:.80-24x=2-x#

#:.-24x+x=2-80#

#-23x=-78#

Multiply both sides by #-1#

#:.23x=78#

#:.color(magenta)(x=78/23# or#color(magenta)(3.391304348#

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#color(magenta)(check:#

#:.2root3(10-3(color(magenta)3.391304348))=root3(2-(color(magenta)3.391304348))#

#:.2root3(10-10.17391304))=root3(2-(3.391304348))#

#:.2root3(-0.17391304))=root3(-1.391304348)#

#:.2 xx -0.558183998=-1.116368004#

#:.color(magenta)(-1.116367997=-1.116368004#