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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Given that [m](a^3 + 3ab^2)/(b^3 + 3a^2b) = 63/62, find the ratio a:b

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Manager  S
Joined: 01 Jun 2019
Posts: 72
Given that [m](a^3 + 3ab^2)/(b^3 + 3a^2b) = 63/62, find the ratio a:b  [#permalink]

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5 00:00

Difficulty:   45% (medium)

Question Stats: 64% (02:45) correct 36% (02:20) wrong based on 25 sessions

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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) $$\frac{6}{7}$$
Intern  B
Joined: 07 Jan 2018
Posts: 39
Re: Given that [m](a^3 + 3ab^2)/(b^3 + 3a^2b) = 63/62, find the ratio a:b  [#permalink]

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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
Intern  B
Joined: 24 Apr 2020
Posts: 4
Re: Given that [m](a^3 + 3ab^2)/(b^3 + 3a^2b) = 63/62, find the ratio a:b  [#permalink]

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Can someone post a step by step solution?
PS Forum Moderator G
Joined: 18 Jan 2020
Posts: 1071
Location: India
GPA: 4
Re: Given that [m](a^3 + 3ab^2)/(b^3 + 3a^2b) = 63/62, find the ratio a:b  [#permalink]

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=> 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.

Posted from my mobile device
Intern  B
Joined: 19 Mar 2020
Posts: 6
Re: Given that [m](a^3 + 3ab^2)/(b^3 + 3a^2b) = 63/62, find the ratio a:b  [#permalink]

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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
Intern  B
Joined: 24 Apr 2020
Posts: 4
Re: Given that [m](a^3 + 3ab^2)/(b^3 + 3a^2b) = 63/62, find the ratio a:b  [#permalink]

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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

3(9+3*4) / 2(4+3*9) = 3*21 / 2*31 = 63/62 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  