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Math Expert V
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Difficulty:   45% (medium)

Question Stats: 66% (01:36) correct 34% (01:32) wrong based on 241 sessions

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Tough and Tricky questions: Algebra.

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

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. $$(x^3 + y^3)(x^2 + xy + y^2)(x - y)$$

Kudos for a correct solution.

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GMAT 1: 700 Q47 V39 Re: x^6 - y^6 =  [#permalink]

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I went with the picking answers method here since I couldn’t immediately simplify in my head:

x^6 – y^6 = (x^3 – y^3)(x^3 + y^3)

I tested the equations that had either (x^3 – y^3) or (x^3 + y^3) in them and went with E since:

(x^2 + xy + y^2)(x – y) simplifies to: x^3 + x^2y + xy^2 – x^2y –xy^2 - y^3, and after canceling: x^3 – y^3

So (x^3 + y^3)((x^2 + xy + y^2)(x – y) = (x^3 + y^3)(x^3 – y^3) = x^6 – x^3y^3 + x^3y^3 - y^6 = x^6 – y^6
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Bunuel wrote:

Tough and Tricky questions: Algebra.

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

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. $$(x^3 + y^3)(x^2 + xy + y^2)(x - y)$$

Kudos for a correct solution.

(x^6-y^6)
= (x^3)^2 - (y^3)^2
= (x^3+y^3)(x^3-y^3)
= (x^3+y^3)(x^2+xy+y^2)(x-y) [ factorizing (x^3-y^3) to get (x^2+xy+y^2)(x-y) ]
E.
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(x^6-y^6)
= (x^3)^2 - (y^3)^2
= (x^3+y^3)(x^3-y^3)
= (x^3+y^3)(x-y)(x^2+Y^2+xy)

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1

$$x^6 - y^6$$

$$= (x^3 + y^3) (x^3 - y^3)$$

$$= (x^3 + y^3)(x-y)(x^2 + xy + y^2)$$
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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: $$(x^3 - y^3)^2 = x^6 - 2x^3y^3 + y^6$$. (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.

The correct answer is (E).

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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GRE 1: Q169 V154 Re: x^6 - y^6 =  [#permalink]

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[quote="Bunuel"]

Tough and Tricky questions: Algebra.

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

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. $$(x^3 + y^3)(x^2 + xy + y^2)(x - y)$$

here the easy way out is to factorize
=(x3+y3)(x3−y3)=(x3+y3)(x3−y3)

=(x3+y3)(x−y)(x2+xy+y2)=(x3+y3)(x−y)(x2+xy+y2)
Here i divided the x^3 -y^3 by x-y as x-y is a factor (0,0)
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Re: x^6 - y^6 =  [#permalink]

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Bunuel wrote:
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.

The correct answer is (E).

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.
$$(8 + 1)(4 + 1)(2 - 1) = 45$$
$$(8 - 1)(8 - 1) = 49$$
$$(4 + 1)(4 + 1)(2 + 1)(2 - 1) = 75$$
$$(16 + 1)(2 + 1)(2 - 1) = 51$$
$$(8 + 1)(4 + 2 + 1)(2 - 1) = 63$$

Bunuel I understand how you sort of reversed engineered the problem - because we should know (X-Y) (X+Y) = X^2-Y^2 without having to foil...but how are you getting (X^3)2- (Y^3)2?
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Nunuboy1994 wrote:
Bunuel wrote:
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.

The correct answer is (E).

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.
$$(8 + 1)(4 + 1)(2 - 1) = 45$$
$$(8 - 1)(8 - 1) = 49$$
$$(4 + 1)(4 + 1)(2 + 1)(2 - 1) = 75$$
$$(16 + 1)(2 + 1)(2 - 1) = 51$$
$$(8 + 1)(4 + 2 + 1)(2 - 1) = 63$$

Bunuel I understand how you sort of reversed engineered the problem - because we should know (X-Y) (X+Y) = X^2-Y^2 without having to foil...but how are you getting (X^3)2- (Y^3)2?

Dear 'Nunuboy1994 please be more specific. Which part of the solution are you talking about. Please highlight!
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Bunuel wrote:
Nunuboy1994 wrote:
Bunuel wrote:
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.

The correct answer is (E).

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.
$$(8 + 1)(4 + 1)(2 - 1) = 45$$
$$(8 - 1)(8 - 1) = 49$$
$$(4 + 1)(4 + 1)(2 + 1)(2 - 1) = 75$$
$$(16 + 1)(2 + 1)(2 - 1) = 51$$
$$(8 + 1)(4 + 2 + 1)(2 - 1) = 63$$

Bunuel I understand how you sort of reversed engineered the problem - because we should know (X-Y) (X+Y) = X^2-Y^2 without having to foil...but how are you getting (X^3)2- (Y^3)2?

Dear 'Nunuboy1994 please be more specific. Which part of the solution are you talking about. Please highlight!

Bunuel

$$x^6 - y^6 = (x^3)2 - (y^3)2 = (x^3 + y^3)(x^3 - y^3)$$ - I don't understand how you got

(x^3)2 - (y^3)2

I get that x^6 - y^6 =(x^3 + y^3)(x^3 - y^3) because (x+y)(x-y) = x^2 - Y^2
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Nunuboy1994 wrote:
Bunuel

$$x^6 - y^6 = (x^3)2 - (y^3)2 = (x^3 + y^3)(x^3 - y^3)$$ - I don't understand how you got

(x^3)2 - (y^3)2

I get that x^6 - y^6 =(x^3 + y^3)(x^3 - y^3) because (x+y)(x-y) = x^2 - Y^2

(x^y)^z = x^(yz), so (x^3)^2 = x^(3*2) = x^6.
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x^6 - y^6 =  [#permalink]

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BEST WAY IS TAKE x= 2 and y= 1
ONLY E WILL GIVE YOU ANSWER
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Re: x^6 - y^6 =  [#permalink]

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Bunuel wrote:

Tough and Tricky questions: Algebra.

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

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. $$(x^3 + y^3)(x^2 + xy + y^2)(x - y)$$

Kudos for a correct solution.

$$x^6 - y^6$$
= $$(x^3 + y^3)*(x^3 - y^3)$$
= $$(x^3 + y^3)*(x - y)*(x^2+xy+y^2)$$

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Re: x^6 - y^6 =  [#permalink]

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Bunuel wrote:
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: $$(x^3 - y^3)^2 = x^6 - 2x^3y^3 + y^6$$. (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.

The correct answer is (E).

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.
$$(8 + 1)(4 + 1)(2 - 1) = 45$$
$$(8 - 1)(8 - 1) = 49$$
$$(4 + 1)(4 + 1)(2 + 1)(2 - 1) = 75$$
$$(16 + 1)(2 + 1)(2 - 1) = 51$$
$$(8 + 1)(4 + 2 + 1)(2 - 1) = 63$$

In this case, I would definitely plug in numbers and do process elimination. B is also wrong because a negative times a negative equal positive. Hope it's clear.
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Re: x^6 - y^6 =  [#permalink]

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Easy question. x^6 - y^6 can also be written as (x^3)^2 - (y^3)^2. This can be expanded as (x^3 - y^3)(x^3 + y^3).

(x^3 - y^3) = (x - y)(x^2 + y^2 + xy)
(x^3 + y^3) = (x + y)(x^2 + y^2 - xy)

Option E is correct as per above formulas.
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