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If ab ≠ 0, a^8 - b^8/(a^4 +b^4)(a^2 + b^2) = ?
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05 Jan 2022, 06:08
05 Jan 2022, 06:08
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If \(ab ≠ 0\), \(\frac{a^8 - b^8}{(a^4 +b^4)(a^2 + b^2) }= ?\)
(A) 1
(B) \(a – b\)
(C) \((a + b)(a – b)\)
(D) \((a^2 + b^2)(a^2 – b^2)\)
(E) \(\frac{a -b}{a+ b}*2√2\)
(A) 1
(B) \(a – b\)
(C) \((a + b)(a – b)\)
(D) \((a^2 + b^2)(a^2 – b^2)\)
(E) \(\frac{a -b}{a+ b}*2√2\)
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Re: If ab ≠ 0, a^8 - b^8/(a^4 +b^4)(a^2 + b^2) = ?
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05 Jan 2022, 06:14
05 Jan 2022, 06:14
1
Carcass wrote:
If \(ab ≠ 0\), \(\frac{a^8 - b^8}{(a^4 +b^4)(a^2 + b^2) }= ?\)
(A) 1
(B) \(a – b\)
(C) \((a + b)(a – b)\)
(D) \((a^2 + b^2)(a^2 – b^2)\)
(E) \(\frac{a -b}{a+ b}*2√2\)
(A) 1
(B) \(a – b\)
(C) \((a + b)(a – b)\)
(D) \((a^2 + b^2)(a^2 – b^2)\)
(E) \(\frac{a -b}{a+ b}*2√2\)
Given: \(\frac{a^8 - b^8}{(a^4 +b^4)(a^2 + b^2) }= ?\)
Since the numerator is a difference of squares we can factor it as follows: \(\frac{a^8 - b^8}{(a^4 +b^4)(a^2 + b^2) }= \frac{(a^4 +b^4)(a^4 -b^4)}{(a^4 +b^4)(a^2 + b^2) }\)
Since the \(a^4 -b^4\) is a difference of squares we can factor it as follows: \(\frac{a^8 - b^8}{(a^4 +b^4)(a^2 + b^2) }= \frac{(a^4 +b^4)(a^2 + b^2)(a^2 - b^2)}{(a^4 +b^4)(a^2 + b^2) }\)
Simplify to get:\(\frac{a^8 - b^8}{(a^4 +b^4)(a^2 + b^2) }= a^2 - b^2 \)
Factor the right side to get: \(\frac{a^8 - b^8}{(a^4 +b^4)(a^2 + b^2) }= (a + b)(a - b)\)
Answer: C
Re: If ab 0, a^8 - b^8/(a^4 +b^4)(a^2 + b^2) = ?
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15 Apr 2024, 05:57
15 Apr 2024, 05:57
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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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