Question 1123069
<pre>{{{((3x^4)/(2y^(-3)))^(-3)}}}

First give the coefficient numbers 3 and 2 the exponent 1:

{{{((3^1x^4)/(2^1y^(-3)))^(-3)}}}

Now that every factor has an exponent, we remove the parentheses,
by multiplying every exponent inside the parentheses by the exponent
outside the parentheses.  [This is the distributive principle of
exponents],

{{{(3^(1*(-3))x^(4*(-3)))/(2^(1*(-3))y^(-3*(-3)))}}}

Simplify:

{{{(3^(-3)x^(-12))/(2^(-3)y^9)}}}

Now use the rule for making negative exponents positive, which is

If there is a negative exponent on the TOP, move the base and
exponent to the BOTTOM and change the sign of the negative
exponent to positive.

and 

If there is a negative exponent on the BOTTOM, move the base and
exponent to the TOP and change the sign of the negative
exponent to positive.

So we move the 3<sup>-3</sup> from the top to the bottom and put 3<sup>3</sup> on the bottom.

We move the x<sup>-12</sup> from the top to the bottom and put x<sup>12</sup> on the bottom.  

And we move the 2<sup>-3</sup> from the bottom to the top and put 2<sup>3</sup> on the top.

But we leave the y<sup>9</sup> on the bottom because it already
has a positive exponent.

{{{(2^3)/(3^3x^12y^9)}}}

We finish by raising 2 to the 3 power and getting 2&#8729;2&#8729;2 = 8,
and raising 3 to the 3 power and getting 3&#8729;3&#8729;3 = 27:

{{{8/(27x^12y^9)}}}

Edwin</pre>