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

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Intern
Joined: 10 Oct 2012
Posts: 25
If (a +1/a)^2=3, find the value of a^3 + 1/a^3  [#permalink]

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Updated on: 13 Nov 2015, 00:26
3
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Difficulty:

75% (hard)

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

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

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.
Renamed the topic
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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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12 Nov 2015, 17:16
4
3
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.
##### General Discussion
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Joined: 29 Jul 2015
Posts: 155
Re: If (a +1/a)^2=3, find the value of a^3 + 1/a^3  [#permalink]

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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

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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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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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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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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

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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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15 Oct 2018, 07:08
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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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