x^6 - y^6 =

(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)


Answer is E.

Answer = E

\(x^6 - y^6\)

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

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

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

Answer: E.

[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)

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

(x^y)^z = x^(yz), so (x^3)^2 = x^(3*2) = x^6.

BEST WAY IS TAKE x= 2 and y= 1
ONLY E WILL GIVE YOU ANSWER

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

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.

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.

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

IMO E

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