Question 288144
Simplify {{{(xy^2z^3)/(x^3y^2z)}}}
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Two ways to do it:

First way:

{{{(xy^2z^3)/(x^3y^2z)}}}

Write each exponential as a product of factors:

{{{(x*y*y*z*z*z)/(x*x*x*y*y*z)}}}

Cancel the x in the top and one x on the bottom:

{{{(cross(x)*y*y*z*z*z)/(cross(x)*x*x*y*y*z)}}}

Cancel the two y's in the top and the two y's on the bottom:

{{{(cross(x)*cross(y*y)*z*z*z)/(cross(x)*x*x*cross(y*y)*z)}}}

Cancel one of the z's in the top and the z on the bottom:

{{{(cross(x)*cross(y*y)*cross(z)*z*z)/(cross(x)*x*x*cross(y*y)*cross(z))}}}

You are left with

{{{(z*z)/(x*x)}}}

which is equivalent to

{{{z^2/x^2}}} choice B.

Second way.

Use the principle:

If there is an exponential in the top and
one in the bottom with the same base, subtract
the exponents "larger minus smaller" and place
the result in the top if the larger exponent
was on top, and on the bottom if the larger
exponent is on the bottom.

{{{(xy^2z^3)/(x^3y^2z)}}}

Write the {{{x}}} in the top as {{{x^1}}}
and the {{{z}}} in the bottom as {{{z^1}}}

{{{(x^1y^2z^3)/(x^3y^2z^1)}}}

Subtract the exponents of {{{x}}}:  3-1 gives 2 and since the
larger exponent is on the bottom we get {{{x^2}}} on the bottom.

The {{{y^2}}} factors cancel out.

Subtract the exponents of {{{z}}}:  3-1 gives 2 and since the
larger exponent is in the top we get {{{z^2}}} in the top, so

we get

{{{z^2/x^2}}} choice B.

The second way is the most efficient way, but the first way
is quite correct also.

Edwin</pre>