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PLs. help me simplifying this: x^3+64y^3/(x^2-4xy+16y^2)(x^2-16y^2)
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holy moly
my ans is.. \[1/x-4y\]
i know im wrong
numerator is the sum of two cubes so it factor as \[x^3+64y^3=(x+4y)(x^2-4xy+16)\]
yah, i got that
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one factor cancels. then cancel the \[x+4y\] and you get \[\frac{1}{x-4y}\]
ok looks like you got these no problem. factor and cancel gets it
you are not wrong. you are right. done
thank you again
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