Question 141488
{{{(x^3-64)/(8x^3+1)}}}.{{{(4x^2-1)/(x^2+4x+16)}}}
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Factoring rules needed:

The difference of two squares factors like this:
{{{FIRST^2-SECOND^2=(FIRST-SECOND)(FIRST+SECOND)}}}

The sum of two cubes factors like this:
To factor {{{FIRST^3+SECOND^2=(FIRST+SECOND)(FIRST^2-FIRST*SECOND+SECOND^2)}}}

The difference of two cubes factors like this:
To factor {{{FIRST^3-SECOND^2=(FIRST-SECOND)(FIRST^2+FIRST*SECOND+SECOND^2)}}}
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We factor {{{x^3-64}}} by writing it as {{{x^3-4^3}}}.
So it's the difference of two cubes with {{{FIRST=x}}} and {{{SECOND=4}}}
So it factors as {{{(x-4)(x^2+x*4+4^2)=(x-4)(x^2+4x+16)}}}

We factor {{{8x^3+1}}} by writing it as {{{(2x)^3+(1)^3}}}.
So it's the sum of two cubes with {{{FIRST=2x}}} and {{{SECOND=1}}}
So it factors as {{{(2x+1)((2x)^2-(2x)*(1)+(1)^2)=(2x+1)(4x^2-2x+1)}}}

We factor {{{4x^2-1}}} by writing it as {{{(2x)^2-(1)^2}}}.
So it's the difference of two squares with {{{FIRST=2x}}} and {{{SECOND=1}}}
So it factors as {{{(2x-1)(2x+1)}}}

So we replace everything by its factorization:

{{{((x-4)(x^2+4x+16))/((2x+1)(4x^2-2x+1))}}}.{{{((2x-1)(2x+1))/(x^2+4x+16)}}} 

We can cancel the {{{(x^2+4x+16)}}}'s

{{{((x-4)(cross(x^2+4x+16)))/((2x+1)(4x^2-2x+1))}}}.{{{((2x-1)(2x+1))/(cross(x^2+4x+16)))}}}

And also we can cancel the {{{(2x+1)}}}'s

{{{((x-4)(cross(x^2+4x+16)))/(cross((2x+1))(4x^2-2x+1))}}}.{{{((2x-1)cross((2x+1)))/(cross(x^2+4x+16)))}}}

That leaves:

{{{((x-4)(2x-1))/(4x^2-2x+1)}}}

It's not necessary, but you can FOIL
out the numerator and get

{{{(2x^2-9x+4)/(4x^2-2x+1)}}}

Edwin</pre>