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If ab 0,
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19 May 2018, 15:20
19 May 2018, 15:20
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Question Stats:
86% (00:56) correct
14% (02:01) wrong based on 50 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 Mar 2026, 04:12
24 Mar 2026, 04:12
Hello from the GRE Prep Club BumpBot!
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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).
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