Question 63567
reduce [64y^3+27z^3] / [4y+3z]
The numerator is the sum of cubes (4y)^3 + (3z)^3 and is 
divisible by 4y+3z leaving a quotient of:

=[(4y)^2-(4y)(3z)+(3z)^2
=16y^2-12yz+9z^2
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Cheers,
Stan H.