Question 157139
{{{(x^2+xy-2y^2)/(x^3+x^2y)}}}*{{{x/(x^2+3xy+2y^2)}}}
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Factor the left top as {{{(x+2y)(x-y)}}}
Factor the left bottom as {{{x^2(x+y)}}}
Factor the right bottom as {{{(x+2y)(x+y)}}}

{{{((x+2y)(x-y))/(x^2(x+y))}}}*{{{x/((x+2y)(x+y))}}}

Indicate multiplication of tops and bottoms, so it
will all be just one fraction:

{{{((x+2y)(x-y)x)/(x^2(x+y)(x+2y)(x+y))}}}

Cancel the {{{(x+2y)}}}'s

{{{(cross((x+2y))(x-y)x)/(x^2(x+y)cross((x+2y))(x+y))}}}

Cancel the {{{x}}} in the top into the {{{x^2}}} on the bottom,
getting {{{x}}} on the bottom:

{{{(cross((x+2y))(x-y)cross(x))/(x^(cross(2))(x+y)cross((x+2y))(x+y))}}}

Now what's left is:

{{{(x-y)/(x(x+y)(x+y))}}}

And {{{(x+y)(x+y)}}} can be written as {{{(x+y)^2}}}

{{{(x-y)/(x(x+y)^2)}}}