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If xy ≠ 0 and x ≠ –y, (x^36 - y^36)/[(x^18 + y^18)(x^9 + y^9)] ?

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If xy ≠ 0 and x ≠ –y, (x^36 - y^36)/[(x^18 + y^18)(x^9 + y^9)] ?  [#permalink]

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17 Jul 2018, 23:50
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Difficulty:

15% (low)

Question Stats:

86% (01:01) correct 14% (02:56) wrong based on 29 sessions

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If xy ≠ 0 and x ≠ –y, $$\frac{x^{36}-y^{36}}{(x^{18}+y^{18})(x^{9}+y^{9})}$$

(A) 1

(B) $$x^2 - y^2$$

(C) $$x^9 - y^9$$

(D) $$x^{18} - y^{18}$$

(E) $$\frac{1}{x^{9}-y^{9}}$$

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Re: If xy ≠ 0 and x ≠ –y, (x^36 - y^36)/[(x^18 + y^18)(x^9 + y^9)] ?  [#permalink]

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18 Jul 2018, 00:04
Bunuel wrote:
If xy ≠ 0 and x ≠ –y, $$\frac{x^{36}-y^{36}}{(x^{18}+y^{18})(x^{9}+y^{9})}$$

(A) 1

(B) $$x^2 - y^2$$

(C) $$x^9 - y^9$$

(D) $$x^{18} - y^{18}$$

(E) $$\frac{1}{x^{9}-y^{9}}$$

As all we're given are equations, we'll work with them.
This is a Precise approach.

Looking at our numerator, we should be reminded of the identity a^2 - b^2 = (a+b)(a-b)
Using this identity, we can write x^36 - y^36 = (x^18 + y^18)(x^18 - y^18)
Canceling out (x^18 + y^18) from numerator and denominator gives (x^18 - y^18)/(x^9 + y^9)
Once again using the same identity, we can write (x^18 - y^18) = (x^9 + y^9)(x^9 - y^9)
Canceling out (x^9 + y^9) from numerator and denominator gives our final answer, (x^9 - y^9)

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Re: If xy ≠ 0 and x ≠ –y, (x^36 - y^36)/[(x^18 + y^18)(x^9 + y^9)] ?  [#permalink]

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18 Jul 2018, 00:18
x^36−y^36 / (x^18 + y^18)*(x^9 + y^9)
((x^18 + y^18) * (x^18 - y^18)) / ((x^18 + y^18)*(x^9 + y^9))
(x^18 - y^18) / (x^9 + y^9)
((x^9 - y^9)*(x^9 + y^9)) / (x^9 + y^9)
(x^9 - y^9)

Hence, C.
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Re: If xy ≠ 0 and x ≠ –y, (x^36 - y^36)/[(x^18 + y^18)(x^9 + y^9)] ?   [#permalink] 18 Jul 2018, 00:18
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