Question 196224
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Note: Josmiceli's solution is not acceptable for this is not supposed to be done by fractional exponents!
Edwin's solution:</b></font>
<pre><font size = 4 color = "indigo"><b>
{{{root(3,24x^5y^3)}}}

Break 24 down into prime factors:

{{{24 = 2*2*2*3}}}

Break {{{x^5}}} down into prime factors:

{{{x^5 = x*x*x*x*x}}}

Break {{{y^3}}} down into prime factors:

{{{y^3=y*y*y}}}

Substituting:

{{{root(3,24x^5y^3)}}}

{{{root(3,2*2*2*3*x*x*x*x*x*y*y*y)}}}

Since the index is 3, use parentheses to
group as many groups of three like factors
as possible under the radical:

{{{root(3,(2*2*2)*3*(x*x*x)*x*x*(y*y*y))}}}

Write each group of three as a cube:

{{{root(3,(2^3)*3*(x^3)*x*x*(y^3))}}}

Take the cube roots of those cubes outside
in front of the radical.  Leave the factors
that did not group under the radical:

{{{2*x*y*root(3,3*x*x)}}}

Write the {{{x*x}}} as {{{x^2}}} under the
radical:

{{{2*x*y*root(3,3*x^2)}}}

{{{2xy*root(3,3x^2)}}}

Edwin</pre>