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# S98-14

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Math Expert
Joined: 02 Sep 2009
Posts: 49320

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16 Sep 2014, 01:52
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Difficulty:

25% (medium)

Question Stats:

74% (00:58) correct 26% (00:43) wrong based on 31 sessions

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$$x^6 - y^6 =$$

A. $$(x^3 + y^3)(x^2 + y^2)(x - y)$$
B. $$(x^3 - y^3)(x^3 - y^3)$$
C. $$(x^2 + y^2)(x^2 + y^2)(x + y)(x - y)$$
D. $$(x^4 + y^4)(x + y)(x - y)$$
E. $$(x^3 + y^3)(x^2 + xy + y^2)(x - y)$$

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 49320

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16 Sep 2014, 01:52
Official Solution:

$$x^6 - y^6 =$$

A. $$(x^3 + y^3)(x^2 + y^2)(x - y)$$
B. $$(x^3 - y^3)(x^3 - y^3)$$
C. $$(x^2 + y^2)(x^2 + y^2)(x + y)(x - y)$$
D. $$(x^4 + y^4)(x + y)(x - y)$$
E. $$(x^3 + y^3)(x^2 + xy + y^2)(x - y)$$

An efficient way to attack this problem is to rephrase the expression in the stem. Since a sixth power is a square, we can look at the difference of sixth powers as the difference of squares, and factor:
$$x^6 - y^6 = (x^3)2 - (y^3)2 = (x^3 + y^3)(x^3 - y^3)$$

This new expression is not among the answer choices, but as a search strategy, we should focus on these factors.

Choice (A) contains $$(x^3 + y^3)$$, so we should determine whether the rest of (A) works out too $$(x^3 - y^3)$$. The terms $$(x^2 + y^2)(x - y)$$ look promising, since they give us $$+x^3$$ and $$-y^3$$, but we get cross-terms that don't cancel: $$(x^2 + y^2)(x - y) = x^3 - x^2y + xy^2 - y^3$$.

Choice (B) is close but not right. $$(x^3 - y^3)(x^3 - y^3) = (x^3 - y^3)^2$$, which also gives us cross-terms that don't cancel: $$(x3 - y3)^2 = x^6 - 2x^3y^3 + y^6$$. (Moreover, the sign is wrong on the $$y^6$$ term.)

Let's skip to choice (E), since we see $$(x^3 + y^3)$$. The remaining terms multiply out as follows:

$$(x^2 + xy + y^2)(x - y) = x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3 = x^3 - y^3$$. This is what we were looking for. We can verify that the other answer choices do not multiply out to $$x^6 - y^6$$ exactly.

Of course, you can also plug simple numbers. Don't pick 0 for either $$x$$ or $$y$$, because too many terms will cancel. Also, don't pick the same number for $$x$$ and $$y$$, because then the result is 0 (and every answer choice gives you 0 as well). But if you pick 2 and 1, you can keep the quantities relatively small and still eliminate wrong answers.

$$2^6 - 1^6 = 64 - 1 = 63$$. Our target is 63.
$$(8 + 1)(4 + 1)(2 - 1) = 45$$
$$(8 - 1)(8 - 1) = 49$$
$$(4 + 1)(4 + 1)(2 + 1)(2 - 1) = 75$$
$$(16 + 1)(2 + 1)(2 - 1) = 51$$
$$(8 + 1)(4 + 2 + 1)(2 - 1) = 63$$

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02 Jul 2018, 01:48
Any further questions like the above?
Math Expert
Joined: 02 Sep 2009
Posts: 49320

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02 Jul 2018, 02:32
Re: S98-14 &nbs [#permalink] 02 Jul 2018, 02:32
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# S98-14

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