Question 174208
<pre><font size = 4 color = "indigo"><b>
 
{{{(matrix(5,1,
"",
4^(-3/2)x^(2/3)y^(-7/4),
"----------------------", "",
2^(3/2) x^(-1/3) y^(3/4)   ))^(2/3)}}}

To remove the big parentheses, multiply
each of the six exponents inside the
parentheses, top and bottom, by the outer
exponent {{{2/3)}}}.

{{{matrix(5,1,
"",
4^((-3/2)(2/3))x^((2/3)(2/3))y^((-7/4)(2/3)),
"----------------------------------", "",
2^((3/2)(2/3)) x^((-1/3)(2/3))y^((3/4)(2/3))   )}}}

{{{matrix(5,1,
"",
4^(-1)x^(4/9)y^(-7/6),
"-------------------", "",
2^(1) x^(-2/9)y^(1/2)   )}}}

Move the factors with negative exponents in
the top to the bottom, changing the signs of
the negative exponents to positive. Also,
move the factors with negative exponents in
the bottom to the top, changing the signs of
the negative exponents to positive.

{{{matrix(5,1,
"",
x^(4/9)x^(2/9),
"-------------------", "",
4^1* y^(7/6)2^1y^(1/2)   )}}}

Add the exponents of x in the top {{{4/9+2/9=6/9=2/3}}}
Add the exponents of y in the bottom {{{7/6+1/2=7/6+3/6=10/6=5/3}}}
Write {{{4^1*2^1}}} as {{{4*2=8}}}


{{{matrix(5,1,
"",
x^(2/3),
"---------", "",
8 y^(5/3)   )}}}

That may be acceptable as the final answer.

However we can rationalize the denominator by
multiplying top and bottom by {{{y^(1/3)}}}

{{{matrix(5,1,
"",
x^(2/3)*y^(1/3),
"---------", "",
8 y^(5/3)*y^(1/3)   )}}}

Write the factors in the top
as radicals. Add exponents on 
the bottom.

{{{matrix(5,1,
"",
root(3,x^2)root(3,y),
"---------", "",
8 y^(6/3)   )}}}

Multiply under the radicals
and simplify the exponent in the
bottom:

{{{matrix(4,1,
"",
root(3,x^2y),
"-------", 
8 y^2   )}}}


Edwin</pre>