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# Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (

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Senior Manager
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Joined: 02 Nov 2018
Posts: 280
Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

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11 Apr 2019, 13:32
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35% (medium)

Question Stats:

72% (02:10) correct 28% (02:24) wrong based on 46 sessions

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Let a and b be relatively prime integers with a > b > 0 and $$\frac{a^3 −b^3}{(a−b)^3}$$ =$$\frac{73}{3}$$ What is a − b?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

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Intern
Joined: 31 Aug 2016
Posts: 5
Location: India
Concentration: Finance, Economics
GMAT 1: 600 Q49 V25
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Re: Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

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11 Apr 2019, 14:25
Can someone explain me how to actually solve this.
I got "C" because I assumed (a-b) should be an integer and multiple of 3 because of 3 in denominator of 73/3

Manager
Joined: 21 Feb 2019
Posts: 125
Location: Italy
Re: Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

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11 Apr 2019, 15:23
2
vedavyas003 wrote:
Can someone explain me how to actually solve this.
I got "C" because I assumed (a-b) should be an integer and multiple of 3 because of 3 in denominator of 73/3

I think the "logic way" is the best here since algebraic method is too long to answer to the question in less than 2 minutes.

Just rewrite it as: $$\frac{(a-b)(a^2 + ab + b^2)}{{(a-b)^3}} = \frac{73}{3}$$.

Taking into account the value of the denominator (3), you should discard all options but C, since the other ones will never give a 3 when simplified with something above the fraction.

However, I'm interested if there is a more methodical approach enough fast to deal with.
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Re: Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

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15 Apr 2019, 18:27
Can someone please post solution to this problem?
Director
Joined: 19 Oct 2018
Posts: 985
Location: India
Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

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15 Apr 2019, 19:34
2
(a-b) (a^2+b^2+ab)/(a-b) (a^2+b^2-2ab)=73/3
1+(3ab/a^2+b^2-2ab)=73/3
ab/(a-b)^2=70/9
as a and b are relatively prime, there will be no common factor between their difference and product
we can further find their values
a=10, b=7
Intern
Joined: 04 Apr 2019
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Re: Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

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16 Apr 2019, 03:15
nick1816 wrote:
(a-b) (a^2+b^2+ab)/(a-b) (a^2+b^2-2ab)=73/3
1+(3ab/a^2+b^2-2ab)=73/3
ab/(a-b)^2=70/9
as a and b are relatively prime, there will be no common factor between their difference and product
we can further find their values
a=10, b=7

Thanks, but how did you get to ab=70?

Posted from my mobile device
Intern
Joined: 18 Jan 2019
Posts: 28
Re: Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

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16 Apr 2019, 05:34
lucajava wrote:
vedavyas003 wrote:
Can someone explain me how to actually solve this.
I got "C" because I assumed (a-b) should be an integer and multiple of 3 because of 3 in denominator of 73/3

I think the "logic way" is the best here since algebraic method is too long to answer to the question in less than 2 minutes.

Just rewrite it as: $$\frac{(a-b)(a^2 + ab + b^2)}{{(a-b)^3}} = \frac{73}{3}$$.

Taking into account the value of the denominator (3), you should discard all options but C, since the other ones will never give a 3 when simplified with something above the fraction.

However, I'm interested if there is a more methodical approach enough fast to deal with.

can you please explain this with more detail?
Manager
Joined: 27 Oct 2017
Posts: 72
Re: Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

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17 Apr 2019, 10:21
nick1816 wrote:
(a-b) (a^2+b^2+ab)/(a-b) (a^2+b^2-2ab)=73/3
1+(3ab/a^2+b^2-2ab)=73/3
ab/(a-b)^2=70/9
as a and b are relatively prime, there will be no common factor between their difference and product
we can further find their values
a=10, b=7

Can you please explain it in more detail?
Director
Joined: 13 Mar 2017
Posts: 728
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

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17 Apr 2019, 10:37
nick1816 wrote:
(a-b) (a^2+b^2+ab)/(a-b) (a^2+b^2-2ab)=73/3
1+(3ab/a^2+b^2-2ab)=73/3
ab/(a-b)^2=70/9
as a and b are relatively prime, there will be no common factor between their difference and product
we can further find their values
a=10, b=7

Your solution is really awesome and well thought of.
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Director
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GPA: 3.8
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Re: Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

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17 Apr 2019, 10:52
4
Let a and b be relatively prime integers with a > b > 0 and $$\frac{a^3 −b^3}{(a−b)^3}$$ =$$\frac{73}{3}$$ What is a − b?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

a and b are relatively prime and a>b>0.

$$\frac{a^3 −b^3}{(a−b)^3}$$ =$$\frac{73}{3}$$
$$\frac{(a-b)(a^2 +ab + b^2)}{(a−b)^3}$$ =$$\frac{73}{3}$$
Since a>b>0 So, a-b=/=0
$$\frac{(a^2 + ab + b^2)}{(a−b)^2}$$ =$$\frac{73}{3}$$
$$\frac{(a^2 + ab + b^2)}{(a^2 - 2ab + b^2)}$$ =$$\frac{73}{3}$$
$$\frac{(a^2 + ab + b^2)}{(a^2 - 2ab + b^2)}-1$$ =$$\frac{73}{3}-1$$
$$\frac{(3ab)}{(a^2 - 2ab + b^2)}$$ =$$\frac{70}{3}$$
$$\frac{(ab)}{(a-b)^2}$$ =$$\frac{70}{9}$$
$$\frac{(ab)}{(a-b)^2}$$ =$$\frac{70x^2}{(3x)^2}$$

so, a-b = 3x and ab = 70x^2
At x =1, a-b = 3, ab = 70 -> (3+b)b= 70 ->b^2 +3b -70 = 0 -> (b+10)(b-7) = 0 -> b= 7, a = 10 , a-b = 3
At x= 2, a-b = 6 , ab = 70*2^2, a= 20, b=14

and so on... But a and b are co-prime and hence a-b =3 , ab = 70, a = 10 & b=7.

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Re: Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

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17 Apr 2019, 11:10
shashankism wrote:
Let a and b be relatively prime integers with a > b > 0 and $$\frac{a^3 −b^3}{(a−b)^3}$$ =$$\frac{73}{3}$$ What is a − b?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

a and b are relatively prime and a>b>0.

$$\frac{a^3 −b^3}{(a−b)^3}$$ =$$\frac{73}{3}$$
$$\frac{(a-b)(a^2 +ab + b^2)}{(a−b)^3}$$ =$$\frac{73}{3}$$
Since a>b>0 So, a-b=/=0
$$\frac{(a^2 + ab + b^2)}{(a−b)^2}$$ =$$\frac{73}{3}$$
$$\frac{(a^2 + ab + b^2)}{(a^2 - 2ab + b^2)}$$ =$$\frac{73}{3}$$
$$\frac{(a^2 + ab + b^2)}{(a^2 - 2ab + b^2)}-1$$ =$$\frac{73}{3}-1$$
$$\frac{(3ab)}{(a^2 - 2ab + b^2)}$$ =$$\frac{70}{3}$$
$$\frac{(ab)}{(a-b)^2}$$ =$$\frac{70}{9}$$
$$\frac{(ab)}{(a-b)^2}$$ =$$\frac{70x^2}{(3x)^2}$$

so, a-b = 3x and ab = 70x^2
At x =1, a-b = 3, ab = 70 -> (3+b)b= 70 ->b^2 +3b -70 = 0 -> (b+10)(b-7) = 0 -> b= 7, a = 10 , a-b = 3
At x= 2, a-b = 6 , ab = 70*2^2, a= 20, b=14

and so on... But a and b are co-prime and hence a-b =3 , ab = 70, a = 10 & b=7.

Thanks a lot..

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Re: Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (   [#permalink] 17 Apr 2019, 11:10
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