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Given that [m](a^3 + 3ab^2)/(b^3 + 3a^2b) = 63/62, find the ratio a:b

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Given that [m](a^3 + 3ab^2)/(b^3 + 3a^2b) = 63/62, find the ratio a:b  [#permalink]

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New post 27 May 2020, 01:18
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Given that \(\frac{a^3 + 3ab^2}{b^3 + 3a^2b} = \frac{63}{62}\), find the ratio a : b

(A) \(\frac{5}{3}\)

(B) \(\frac{3}{5}\)

(C) \(\frac{2}{3}\)

(D) \(\frac{3}{2}\)

(E) \(\frac{6}{7}\)
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Re: Given that [m](a^3 + 3ab^2)/(b^3 + 3a^2b) = 63/62, find the ratio a:b  [#permalink]

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New post 27 May 2020, 02:12
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karan12345 wrote:
Given that \(\frac{a^3 + 3ab^2}{b^3 + 3a^2b} = \frac{63}{62}\), find the ratio a : b

(A) \(\frac{5}{3}\)

(B) \(\frac{3}{5}\)

(C) \(\frac{2}{3}\)

(D) \(\frac{3}{2}\)

(E) \(\frac{6}{7}\)


\(a^3 + 3ab^2 / b^3 + 3a^2 b\) =\(63/62\)
By componendo and dividendo,
\((a+b)^3 / (a-b)^3\) = \(125/1\)
Taking cube root:
\((a+b)/ (a-b)\) = \(5/1\)
again applying componendo dividendo
\(2a/2b = 6/4\)
\(a/b = 3/2\)
Ans. D
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Re: Given that [m](a^3 + 3ab^2)/(b^3 + 3a^2b) = 63/62, find the ratio a:b  [#permalink]

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New post 29 May 2020, 09:36
Can someone post a step by step solution?
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Re: Given that [m](a^3 + 3ab^2)/(b^3 + 3a^2b) = 63/62, find the ratio a:b  [#permalink]

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New post 29 May 2020, 10:37
=> A^3+3AB^2/B^3+3A^2B = 63/62
=> Using (A+B)^3 property
=> A^3+3AB^2+B^3+3A^2B-B^3-3A^2B/B^3+3A^2B = 63/62
=> (A+B)^3/B^3+3A^2B - (B^3+3A^2B)/B^3+3A^2B = 63/62
=> (A+B)^3/B^3+3A^2B -1 = 63/62
=> (A+B)^3/B^3+3A^2B =(63/62)+1
=> (A+B)^3/B^3+3A^2B = 125/62
=> Comparing Both sides
=> (A+B)^3 = 125 => {A+B} = 5
=> A+B = 2+3/3+2
=> A and B can be 3:2 or 2:3
=> now put value of A and B in equation (A+B)^3/B^3+3A^2B = 125/62
=> you will get A=3 and B=2 satisfying the equation.

Hence Answer is D

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Re: Given that [m](a^3 + 3ab^2)/(b^3 + 3a^2b) = 63/62, find the ratio a:b  [#permalink]

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New post 29 May 2020, 11:56
bidskamikaze wrote:
karan12345 wrote:
Given that \(\frac{a^3 + 3ab^2}{b^3 + 3a^2b} = \frac{63}{62}\), find the ratio a : b

(A) \(\frac{5}{3}\)

(B) \(\frac{3}{5}\)

(C) \(\frac{2}{3}\)

(D) \(\frac{3}{2}\)

(E) \(\frac{6}{7}\)


\(a^3 + 3ab^2 / b^3 + 3a^2 b\) =\(63/62\)
By componendo and dividendo,
\((a+b)^3 / (a-b)^3\) = \(125/1\)
Taking cube root:
\((a+b)/ (a-b)\) = \(5/1\)
again applying componendo dividendo
\(2a/2b = 6/4\)
\(a/b = 3/2\)
Ans. D
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Re: Given that [m](a^3 + 3ab^2)/(b^3 + 3a^2b) = 63/62, find the ratio a:b  [#permalink]

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New post 01 Jun 2020, 14:35
1
I feel like an easier method may be guess and check or property of factors
63 = 3*7*3
62=2*31
a(a^2 + 3b^2) / b(b^2+3a^2)

Therefore A needs to have a factor of 3 and B a factor of 2

Therefore answer is 3/2, additionally can guess and check to confirm
3(9+3*4) / 2(4+3*9) = 3*21 / 2*31 = 63/62
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Re: Given that [m](a^3 + 3ab^2)/(b^3 + 3a^2b) = 63/62, find the ratio a:b   [#permalink] 01 Jun 2020, 14:35

Given that [m](a^3 + 3ab^2)/(b^3 + 3a^2b) = 63/62, find the ratio a:b

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