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Given that a^{-5}+b^{-5}=0 and $a b \neq 0
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17 Aug 2026, 02:56
17 Aug 2026, 02:56
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Expert Reply
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Question Stats:
55% (01:58) correct
44% (02:25) wrong based on 9 sessions
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Given that \( a^{-5}+b^{-5}=0\) and \( a b \neq 0\) , what is the value of \(\frac{a^2+a b+b^2}{a^2-a b+b^2}\) ?
(A) — 1
(B) 0
(C) \(\frac{1}{3}\)
(D) 1
(E) 3
(A) — 1
(B) 0
(C) \(\frac{1}{3}\)
(D) 1
(E) 3
- Part of the project: GRE Daily Quant Challenge ALL 4 CORE QUANT AREAS
- Also replying to the unanswered questions
Re: Given that a^{-5}+b^{-5}=0 and $a b \neq 0
[#permalink]
18 Aug 2026, 03:32
18 Aug 2026, 03:32
Expert Reply
Step 1: Simplify the given relationship
Start with the equation:
\(a^{-5}+b^{-5}=0\)
Rewrite using positive exponents:
\(\frac{1}{a^5}+\frac{1}{b^5}=0\)
Subtract \(\frac{1}{b^5}\) from both sides:
\(\frac{1}{a^5}=-\frac{1}{b^5}\)
Invert both sides:
\(a^5=-b^5\)
Taking the fifth root of both sides gives:
\( a=-b \quad \text { or } \quad b=-a\)
Step 2: Substitute $b=-a$ into the target expression
We want to evaluate:
\(\frac{a^2+a b+b^2}{a^2-a b+b^2}\)
Substitute b=-a :
Numerator:
\(a^2+a(-a)+(-a)^2=a^2-a^2+a^2=a^2\)
Denominator:
\(a^2-a(-a)+(-a)^2=a^2+a^2+a^2=3 a^2\)
Step 3: Simplify the fraction
\(\frac{a^2}{3 a^2}=\frac{1}{3}\)
Correct Answer: (C) \(\frac{1}{3}\)
Start with the equation:
\(a^{-5}+b^{-5}=0\)
Rewrite using positive exponents:
\(\frac{1}{a^5}+\frac{1}{b^5}=0\)
Subtract \(\frac{1}{b^5}\) from both sides:
\(\frac{1}{a^5}=-\frac{1}{b^5}\)
Invert both sides:
\(a^5=-b^5\)
Taking the fifth root of both sides gives:
\( a=-b \quad \text { or } \quad b=-a\)
Step 2: Substitute $b=-a$ into the target expression
We want to evaluate:
\(\frac{a^2+a b+b^2}{a^2-a b+b^2}\)
Substitute b=-a :
Numerator:
\(a^2+a(-a)+(-a)^2=a^2-a^2+a^2=a^2\)
Denominator:
\(a^2-a(-a)+(-a)^2=a^2+a^2+a^2=3 a^2\)
Step 3: Simplify the fraction
\(\frac{a^2}{3 a^2}=\frac{1}{3}\)
Correct Answer: (C) \(\frac{1}{3}\)
gmatclubot
Re: Given that a^{-5}+b^{-5}=0 and $a b \neq 0 [#permalink]
18 Aug 2026, 03:32
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