Question 217793
{{{2*root(3,3)+4*root(3,24)-root(3,81)}}}
First we simplify each cube root. To simplify cube roots, try to find perfect cube factors (other than 1) in the radicand (the number inside the radical).<br>
The first radicand has no perfect cube factors. The second and third radicands, however, do have perfect cube factors: 8 and 27 respectively. We can then rewrite them:
{{{2*root(3,3)+4*root(3,8*3)-root(3,27*3)}}}
Now we can use the property of roots, {{{root(a, x*y) = root(a, x)*root(a,y)}}}, 
to separate the perfect cube factors into their own cube roots:
{{{2*root(3,3)+4*root(3,8)*root(3,3)-root(3,27)*root(3,3)}}}
Since {{{root(3,8) = 2}}} and {{{root(3, 27) = 3}}} we have:
{{{2*root(3,3)+4*2*root(3,3)-3*root(3,3)}}}
Simplifying the middle term:
{{{2*root(3,3)+8*root(3,3)-3*root(3,3)}}}
Now we have simplified the individual cube roots. Next we try to simplify the entire expression. Can we add and subtract these terms? Answer: yes. They are like terms. Just like 2x+8x-3x are like terms so are these cube roots. If you have trouble seeing this, try using a variable. Let {{{x=root(3,3)}}}. Then our equation becomes 2x+8x-3x! (On the other hand, {{{2*root(3,4)+8*root(3,6)}}} are not like terms because the radicands are different. And {{{2*root(3, 5) + 8*root(4,5)}}} are not like terms because they are different roots.)<br>
{{{2*root(3,3)+8*root(3,3)-3*root(3,3) = (2+8-3)*root(3,3) = 7*root(3,3)}}}
Since no more simplifying can be done, we are finished.