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If ab 0,
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19 May 2018, 15:20
19 May 2018, 15:20
Expert Reply
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Question Stats:
88% (00:57) correct
11% (01:12) wrong based on 45 sessions
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If \(ab \neq 0\), \(\frac{a^8-b^8}{(a^4+b^4)(a^2+b^2)}\)
(A) \(1\)
(B) \(a - b\)
(C) \((a + b)(a - b)\)
(D) \((a^2 + b^2)(a^2 - b^2)\)
(E) \(\frac{a-b}{a+b}\)
(A) \(1\)
(B) \(a - b\)
(C) \((a + b)(a - b)\)
(D) \((a^2 + b^2)(a^2 - b^2)\)
(E) \(\frac{a-b}{a+b}\)
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Re: If ab 0,
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30 Apr 2019, 16:17
30 Apr 2019, 16:17
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sandy wrote:
If \(ab \neq 0\), \(\frac{a^8-b^8}{(a^4+b^4)(a^2+b^2)}\)
(A) \(1\)
(B) \(a - b\)
(C) \((a + b)(a - b)\)
(D) \((a^2 + b^2)(a^2 - b^2)\)
(E) \(\frac{a-b}{a+b}\)
(A) \(1\)
(B) \(a - b\)
(C) \((a + b)(a - b)\)
(D) \((a^2 + b^2)(a^2 - b^2)\)
(E) \(\frac{a-b}{a+b}\)
Key concept: difference of squares.
\(\frac{a^8-b^8}{(a^4+b^4)(a^2+b^2)}=\frac{(a^4+b^4)(a^4-b^4)}{(a^4+b^4)(a^2+b^2)}\)
\(=\frac{(a^4+b^4)(a^2+b^2)(a^2-b^2)}{(a^4+b^4)(a^2+b^2)}\)
\(=a^2-b^2\)
\(=(a+b)(a-b)\)
Answer: C
Cheers,
Brent
General Discussion
Re: If ab 0,
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24 Jan 2024, 12:51
24 Jan 2024, 12:51
Hello from the GRE Prep Club BumpBot!
Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
