Question 1085563
<pre>
{{{(2*root(3,4))*(3*root(3,16))}}}

Take away the parentheses

{{{2*root(3,4)*3*root(3,16)}}}

Rearrange the factors

{{{2*3*root(3,4)*root(3,16)}}}

Since  2•3 = 6, we have:

{{{6*root(3,4)*root(3,16)}}}

Those roots have the same index, so
we can multiply the numbers under
them an just have one cube root:

{{{6*root(3,4*16)}}}

And since 4•16 = 64,

{{{6*root(3,64)}}}

Maybe you already know that the cube root of 64 is 4, and
that you'd have 6•4 which is 24. But if you don't you can 
do this instead:

Break 64 down like this:
64 = 4•4•4 = 4<sup>3</sup>, so we have:

{{{6*root(3,4^3)}}}

When the index of the root and the exponent
of the radicand are the same, the cube root
cancels out the cube:

{{{6*root(cross(3),4^cross(3))}}}

and all that's left is the number under the
radical:

{{{6*4)}}}

And that's just

{{{24}}}

Edwin</pre>