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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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Math Expert
Joined: 02 Sep 2009
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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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08 Oct 2017, 05:32
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Difficulty:

35% (medium)

Question Stats:

66% (00:55) correct 34% (01:11) wrong based on 134 sessions

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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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Joined: 19 Sep 2016
Posts: 35
Re: What is a–b? (1) a^3 – b^3 = 117 (2) a^2 + ab + b^2 = 39  [#permalink]

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08 Oct 2017, 06:31
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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08 Oct 2017, 07:17
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)

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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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09 Sep 2018, 12:06
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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Joined: 21 Aug 2018
Posts: 6
Re: What is a–b? (1) a^3 – b^3 = 117 (2) a^2 + ab + b^2 = 39  [#permalink]

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20 Oct 2018, 19:29
Bunuel wrote:
What is a–b?

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

Hi Bunuel,

Do you have a straight forward way to solve the two statements combined?

Many thanks
Re: What is a–b? (1) a^3 – b^3 = 117 (2) a^2 + ab + b^2 = 39 &nbs [#permalink] 20 Oct 2018, 19:29
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