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What is a–b? (1) a^3 – b^3 = 117 (2) a^2 + ab + b^2 = 39

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What is a–b? (1) a^3 – b^3 = 117 (2) a^2 + ab + b^2 = 39  [#permalink]

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New post 08 Oct 2017, 06:32
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

68% (01:12) correct 32% (01:25) wrong based on 79 sessions

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Re: What is a–b? (1) a^3 – b^3 = 117 (2) a^2 + ab + b^2 = 39  [#permalink]

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New post 08 Oct 2017, 07:31
OA-C, statement 1 can be rewritten as (a-b)^3+3ab(a-b)=117. Which can be further simplified as (a-b)[(a-b)^2+3ab]=117. So (a-b) is dependant on value of 'ab' and hence insufficient.
Statement 2, can be simplified as (a-b)^2+2ab+ab=39, so it becomes
(a-b)^2+3ab=39. Since value again dependant on value of 'ab', so again insufficient.
St 1+ st2 - (a-b)=117/39 and further (a-b)=3. So my answer is C.

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Re: What is a–b? (1) a^3 – b^3 = 117 (2) a^2 + ab + b^2 = 39  [#permalink]

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New post 08 Oct 2017, 08:17
a^3-b^3 = (a-b)(a^2+ab+b^2)

1) a^3-b^3 = 117, Insufficient
2) (a^2+ab+b^2), Insufficient

Taking both option together we can derive the value of (a-b)

Answer C
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Re: What is a–b? (1) a^3 – b^3 = 117 (2) a^2 + ab + b^2 = 39  [#permalink]

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New post 09 Sep 2018, 13:06
Bunuel wrote:
What is a–b?

(1) a^3 – b^3 = 117
(2) a^2 + ab + b^2 = 39

\(? = a - b\)

\(\left( 1 \right)\,\,{a^3} - {b^3} = 117\,\,\,\left\{ \begin{gathered}
\,\,{\text{Take}}\,\,\,\left( {a;b} \right) = \left( {\sqrt[{3\,}]{{117}}\,\,;\,\,0} \right)\,\,\,\,\, \Rightarrow \,\,\,\,? = \sqrt[{3\,}]{{117}}\,\,\, \hfill \\
\,\,{\text{Take}}\,\,\,\left( {a;b} \right) = \left( {1\,\,;\,\, - \sqrt[{3\,}]{{116}}} \right)\,\,\,\,\, \Rightarrow \,\,\,\,? = 1 + \sqrt[{3\,}]{{116}}\,\,\, \ne \sqrt[{3\,}]{{117}}\,\, \hfill \\
\end{gathered} \right.\)


\(\left( 2 \right)\,\,{a^2} + ab + {b^2} = 39\)

\(\left\{ \begin{gathered}
\,\,{\text{Take}}\,\,\,\left( {a;b} \right) = \left( {\sqrt {39} \,\,;\,\,0} \right)\,\,\,\,\, \Rightarrow \,\,\,\,? = \sqrt {39} \,\,\, \hfill \\
\,\,{\text{Take}}\,\,\,\left( {a;b} \right) = \left( {0\,\,;\,\,\sqrt {39} } \right)\,\,\,\,\, \Rightarrow \,\,\,\,? = - \sqrt {39} \,\,\, \ne \sqrt {39} \,\, \hfill \\
\end{gathered} \right.\)


\(\left( {1 + 2} \right)\,\,\,\left\{ \begin{gathered}
\,\,\left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) = {a^3} - {b^3} = 117 \hfill \\
\,\,{a^2} + ab + {b^2} = 39 \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,39\left( {a - b} \right) = 117\,\,\,\,\, \Rightarrow \,\,? = a - b\,\,\,{\text{unique}}\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
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Fabio Skilnik :: http://www.GMATH.net (Math for the GMAT)
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Re: What is a–b? (1) a^3 – b^3 = 117 (2) a^2 + ab + b^2 = 39 &nbs [#permalink] 09 Sep 2018, 13:06
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