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# If x + 2/x = 3 what is the value of x^3 + 8/x^3 = ?

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Intern
Joined: 04 Aug 2010
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If x + 2/x = 3 what is the value of x^3 + 8/x^3 = ? [#permalink]

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Updated on: 28 Nov 2017, 05:09
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Question Stats:

83% (01:02) correct 17% (02:13) wrong based on 252 sessions

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If $$x + \frac{2}{x} = 3$$ what is the value of $$x^3 + \frac{8}{x^3} =$$ ?

A) 1
B) 8
C) 9
D) 16
E) 18

Originally posted by newmoon on 15 Dec 2011, 09:07.
Last edited by Bunuel on 28 Nov 2017, 05:09, edited 3 times in total.
Edited the question.
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Joined: 21 Nov 2011
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Re: If x + 2/x = 3 what is the value of x^3 + 8/x^3 = ? [#permalink]

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15 Dec 2011, 10:06
2
Explanation:
x + 2/x = 3
(x + 2/x)^2 = 9
x^2 + (2/x)^2 + 4 = 9 ====> x^2 + 4/(x^2) = 5
X^3+8/(X^3) = x^3 + (2/x)^3 = (x + 2/x)(x^2+4/(x^2) - 2) = 3 * (5-2) = 9

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Re: If x + 2/x = 3 what is the value of x^3 + 8/x^3 = ? [#permalink]

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19 Dec 2011, 02:14
1
Instead of going the algebraic way , I think the quickest way is the method of substitution. Just replace x=2 in both the cases , and the answer comes as 9 is 20 secs.
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Re: If x + 2/x = 3 what is the value of x^3 + 8/x^3 = ? [#permalink]

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18 Oct 2015, 23:58
1
newmoon wrote:
If x + 2/x = 3 what is the value of x^3 + 8/x^3 = ?

A) 1
B) 8
C) 9
D)16

$$x + 2/x = 3$$
Or $$x^2 + 2 = 3x$$
$$x^2 - 3x +2 = 0$$

The values of x satisfying this equation are 1 and 2
When we put these values in $$x^3 + 8/x^3$$,
we get the answer = 9
Option C
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Re: If x + 2/x = 3 what is the value of x^3 + 8/x^3 = ? [#permalink]

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02 Apr 2016, 01:48
2
newmoon wrote:
If x + 2/x = 3 what is the value of x^3 + 8/x^3 = ?

A) 1
B) 8
C) 9
D)16

Hi,

two other ways to do it..

1) Exploiting the choices..
whenever we have a^3+b^3, it is div by a+b..
and a^3-b^3 is div by a-b..

here
$$x^3 + \frac{8}{x^3} = x^3 + (\frac{2}{x})^3$$ should be div by$$x+\frac{2}{x}$$, which is 3..
so $$x^3 + \frac{8}{x^3}$$ should be multiple of 3..
so 9 should be the answer, as no other choice is miltiple of 3

2) formula
$$a^3+b^3= (a+b)^3-3ab(a+b)$$..
$$x^3 + \frac{8}{x^3} = x^3 + (\frac{2}{x})^3$$= $$(x+\frac{2}{x})^3-3(x*\frac{2}{x}*( x+\frac{2}{x}))$$..
=>$$3^3-3*2*3=27-18=9$$
ans C
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Re: If x + 2/x = 3 what is the value of x^3 + 8/x^3 = ? [#permalink]

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02 Apr 2016, 08:38
ArunpriyanJ wrote:
thisiszico2006 wrote:
Instead of going the algebraic way , I think the quickest way is the method of substitution. Just replace x=2 in both the cases , and the answer comes as 9 is 20 secs.

How did u come to a conclusion to substitute 2? why not other nos?

Hi ArunpriyanJ,

The post that you have quoted is from 2011 and I am sure the poster will not reply so let me pitch in
The person might have started picking numbers randomly from 1 and luckily 2 is the number that fits here.
This is just a brute force method and there is no set pattern here for number picking.
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Re: If x + 2/x = 3 what is the value of x^3 + 8/x^3 = ? [#permalink]

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17 May 2016, 02:41
From the from divisibility rule a^n +b^n is divisible by a+b for all the odd values of n.
so x^3 + 8/x^3 is divisible by x+2/3 for all odd values of n

here n being 3, it should be divisible by 3. Hence answer is option C
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Re: If x + 2/x = 3 what is the value of x^3 + 8/x^3 = ? [#permalink]

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29 Nov 2017, 18:14
newmoon wrote:
If $$x + \frac{2}{x} = 3$$ what is the value of $$x^3 + \frac{8}{x^3} =$$ ?

A) 1
B) 8
C) 9
D) 16
E) 18

We can use the first equation to determine possible values for x. Multiplying the first equation by x we have:

x^2 + 2 = 3x

x^2 - 3x + 2 = 0

(x - 1)(x - 2) = 0

So when x is 1 we have 1 + 8 = 9 and when x is 2 we have 8 + 1 = 9.

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Re: If x + 2/x = 3 what is the value of x^3 + 8/x^3 = ? [#permalink]

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29 Nov 2017, 22:46
newmoon wrote:
If $$x + \frac{2}{x} = 3$$ what is the value of $$x^3 + \frac{8}{x^3} =$$ ?

A) 1
B) 8
C) 9
D) 16
E) 18

x+2/x = 3
x^2 -3x +2 = 0
x = 2,1

If x = 2 , x^3 + 8/x^3 = 2^3 + 8/2^3 = 9
If x = 1, x^3 + 8/x^3 = 1^3 + 8/1^3 = 9

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Joined: 12 Oct 2017
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Re: If x + 2/x = 3 what is the value of x^3 + 8/x^3 = ? [#permalink]

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02 Dec 2017, 00:32
x^3 + 8/x^3 = (x+2/x)^3- 3*x*(2/x)*(x+2/x) = 3^3 - 3*2*3 = 9.
Re: If x + 2/x = 3 what is the value of x^3 + 8/x^3 = ?   [#permalink] 02 Dec 2017, 00:32
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