If I did not recognize that the numerator was the difference of cubes and that I could factor out (a-b)^2 from numerator and denominator, could this be solved using polynomial division?

a^3 – 3a^2b + 3ab^2 – b^3
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(a – b )(a – b)

If I did not recognize that the numerator was the difference of cubes and that I could factor out (a-b)^2 from numerator and denominator, could this be solved using polynomial division?

It could! But if you recognize a shortcut, why do the long way? 🙂

(Just to make sure nobody gets confused, the numerator here isn’t a difference of cubes; it’s actually a regular old binomial cube. That numerator factors to $(a-b)^3$ . So what you really have in this question is $\dfrac{(a-b)^3}{(a-b)^2}$ which of course simplifies to plain old $a-b$ .)

Comments (2)

Doh! Thank you, Mike …. but what if I did NOT recognize that fact? I tried to use polynomial division and got all caught up into knots. Is that a possible way to solve (since I did NOT recognize the full factorization, though could guess that probably both had at least a-b as a factor in common)?

Well, you didn’t submit answer choices with this question but I don’t see how they could have not been in terms of $a$ and $b$ , so another faster and less thorny way to go would have been to plug in. Say $a=2$ and $b=3$ and evaluate the original expression and the answer choices.

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If I did not recognize that the numerator was the difference of cubes and that I could factor out (a-b)^2 from numerator and denominator, could this be solved using polynomial division?

Comments (2)

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