Question 176864
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{{{(a^2b)/(a^5b^3)}}}

First method: breaking everything into prime factors:

Write {{{a^2}}} as {{{a*a}}}
Write {{{a^5}}} as {{{a*a*a*a*a}}}
Write {{{b^3}}} as {{{b*b*b}}}

Then we have:

{{{(a*a*b)/(a*a*a*a*a*b*b*b)}}}

Put a {{{1}}} factor on the top and bottom:

{{{(1*a*a*b)/(1*a*a*a*a*a*b*b*b)}}}

Then we cancel the two {{{a}}}'s on top with two of
the {{{a}}}'s on the bottom:

{{{(1*cross(a)cross(a)*b)/(1*cross(a)cross(a)*a*a*a*b*b*b)}}}

Then we cancel the {{{b}}} in the top into one of the
{{{b}}}'s in the bottom:

{{{(1*cross(a)cross(a)cross(b))/(1*cross(a)cross(a)a*a*a*cross(b)b*b)}}}

So all that didn't cancel out is

{{{1/(a*a*a*b*b)}}}

Then write the {{{a*a*a}}} as {{{a^3}}} and {{{b*b}}} as {{{b^2}}}

{{{1/(a^3b^2)}}}

That way will always work, but here is another method:

-------------

Second method:

{{{(a^2b)/(a^5b^3)}}}

Put a 1 coefficient on top and bottom.
Also put a 1 exponent on the {{{b}}} on top:

{{{(1*a^2b^1)/(1*a^5b^3)}}}

Now subtract the exponents of {{{a}}}
That is subtract them "largest minus smaller"
and put the result where the larger exponent
was, that is, either the numerator or
denominator.

In this case the larger exponents are both in the
bottom, 

so we subtract exponents of {{{a}}},
5-2 getting 3, so we put {{{a^3}}} on the
bottome because the larger exponent 5 was
on the bottom.   

we also subtract exponents of {{{b}}},
3-1 getting 2, so we put {{{b^2}}} on the
bottom because the larger exponent 3 was
on the bottom.

So we have:

{{{(1)/(1*a^3b^2)}}}

or

{{{(1)/(a^3b^2)}}}  

Edwin</pre>