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If (a +1/a)^2=3, find the value of a^3 + 1/a^3

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If (a +1/a)^2=3, find the value of a^3 + 1/a^3  [#permalink]

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New post Updated on: 13 Nov 2015, 00:26
3
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Question Stats:

53% (01:59) correct 47% (02:28) wrong based on 235 sessions

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If \((a +\frac{1}{a})^2=3\), find the value of \(a^3 + \frac{1}{a^3}^\)

A. 0
B. 1
C. \(2+\sqrt{3}\)
D. \(\sqrt{3}\)
E. not enough information


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Originally posted by mmelendez on 12 Nov 2015, 14:50.
Last edited by Bunuel on 13 Nov 2015, 00:26, edited 2 times in total.
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Re: If (a +1/a)^2=3, find the value of a^3 + 1/a^3  [#permalink]

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New post 12 Nov 2015, 17:16
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mmelendez wrote:
if \((a +\frac{1}{a})^2=3\), find the value of \((a^3 + \frac{1}{a^3})\)

A 0
B 1
C \(2+\sqrt{3}\)
D \(\sqrt{3}\)
E not enough information


Kudos if you like this question, and the explanation I will give in 48 hours after discussion


\((a+\frac{1}{a})^2 = 3\) ---> \((a+\frac{1}{a}) = 3^{0.5}\) ...(1)

Now, \((a^3 + \frac{1}{a^3})\) = \((a+\frac{1}{a})^3 - 3*(a+\frac{1}{a})\) ....[ as \((a+b)^3 = a^3+b^3+3ab(a+b)\)]

Thus, \((a^3 + \frac{1}{a^3})\) = \((a+\frac{1}{a})^3 - 3*(a+\frac{1}{a})\) = \(3^{1.5}-3*3^{0.5}\) = 0.

A is the correct answer.
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Re: If (a +1/a)^2=3, find the value of a^3 + 1/a^3  [#permalink]

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New post 13 Nov 2015, 03:21
1
2
mmelendez wrote:
If \((a +\frac{1}{a})^2=3\), find the value of \(a^3 + \frac{1}{a^3}^\)

A. 0
B. 1
C. \(2+\sqrt{3}\)
D. \(\sqrt{3}\)
E. not enough information


Kudos if you like this question, and the explanation I will give in 48 hours after discussion


\(a^3 +b^3\) = \((a+b)(a^2-ab+b^2)\)

\(a^3 + \frac{1}{a^3}^\) = \((a+\frac{1}{a})(a^2 - 1 +\frac{1}{a^2})\) ........(1)

It is given that
\((a +\frac{1}{a})^2=3\)

or \(a +\frac{1}{a}=\sqrt{3}\) ........(2)

Also,
\((a +\frac{1}{a})^2=3\)
or \(a^2+\frac{1}{a^2} +2 =3\)
or \(a^2+\frac{1}{a^2} =1\) ..........(3)

Putting 2 and 3 in 1, we have
\(a^3 + \frac{1}{a^3}^\) = (\(\sqrt{3}\))(1-1) = 0

Answer:- A
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Re: If (a +1/a)^2=3, find the value of a^3 + 1/a^3  [#permalink]

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New post 17 Sep 2016, 04:27
Engr2012 wrote:

\((a+\frac{1}{a})^2 = 3\) ---> \((a+\frac{1}{a}) = 3^{0.5}\) ...(1)

Even though it would not change the answer, we cannot do this manipulation because here you have neglected 1 possible value of \(a+\frac{1}{a}\)

\((a+\frac{1}{a})^2 = 3 => (a+\frac{1}{a})^2 - 3 = 0\)

=> \((a+\frac{1}{a} - \sqrt{3})*(a+\frac{1}{a} + \sqrt{3}) = 0\)

=> \(a+\frac{1}{a} = \sqrt{3}\) OR \(a+\frac{1}{a} = -\sqrt{3}\)
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If (a +1/a)^2=3, find the value of a^3 + 1/a^3  [#permalink]

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New post 17 Sep 2016, 06:41
umg wrote:
Engr2012 wrote:

\((a+\frac{1}{a})^2 = 3\) ---> \((a+\frac{1}{a}) = 3^{0.5}\) ...(1)

Even though it would not change the answer, we cannot do this manipulation because here you have neglected 1 possible value of \(a+\frac{1}{a}\)

\((a+\frac{1}{a})^2 = 3 => (a+\frac{1}{a})^2 - 3 = 0\)

=> \((a+\frac{1}{a} - \sqrt{3})*(a+\frac{1}{a} + \sqrt{3}) = 0\)

=> \(a+\frac{1}{a} = \sqrt{3}\) OR \(a+\frac{1}{a} = -\sqrt{3}\)


Sometimes for the sake of discussion (and saving on time) on this forum, experts will skip a few steps. But you are correct that the original equation will have 2 solutions as it is a quadratic one. There was no "manipulation" involved.
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If (a +1/a)^2=3, find the value of a^3 + 1/a^3  [#permalink]

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New post 26 Jul 2017, 06:30
mmelendez wrote:
If \((a +\frac{1}{a})^2=3\), find the value of \(a^3 + \frac{1}{a^3}^\)

A. 0
B. 1
C. \(2+\sqrt{3}\)
D. \(\sqrt{3}\)
E. not enough information


Kudos if you like this question, and the explanation I will give in 48 hours after discussion


\((a +\frac{1}{a})^2=3\)
\(a +\frac{1}{a} = \sqrt{3}\)

a^3 + 1/a^3 = (a+1/a) (a^2 -a*(1/a) + 1/a^2) = (a+1/a) (a^2 - 1 + 1/a^2) = (a+1/a) ((a+1/a)^2 - 2 -1)
= \((\sqrt{3}(3-3))\) = 0

Answer A
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Re: If (a +1/a)^2=3, find the value of a^3 + 1/a^3  [#permalink]

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Re: If (a +1/a)^2=3, find the value of a^3 + 1/a^3   [#permalink] 15 Oct 2018, 07:08
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