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

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Bunuel
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If (a + 1/a)^2 = 3, find the value of a^3 + 1/a^3 [#permalink] New post   04 Dec 2019, 01:41
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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. \(\sqrt{3}\)

D. \(2+\sqrt{3}\)

E. Not enough information

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A
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Re: If (a + 1/a)^2 = 3, find the value of a^3 + 1/a^3 [#permalink] New post   04 Dec 2019, 02:09
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using (a + b)^3 = a^3 + b^3 + 3ab(a + b)

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

Now substituting the values from question:
a^3+(1/a)^3= (a+1/a)^3-3*a*1/a(a+1/a)
=\(\sqrt{3}\)^3-3*\(\sqrt{3}\)
=3\(\sqrt{3}\)-3\(\sqrt{3}\)
=0

Answer: A
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Re: If (a + 1/a)^2 = 3, find the value of a^3 + 1/a^3 [#permalink] New post   04 Dec 2019, 06:11
Given : \((a+\frac{1}{a})^2 = 3\)

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

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

Lets just bother about LHS for now, Expanding LHS

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

\(a^3 + \frac{1}{a^3 } + 3a + \frac{3}{a}\)

Taking 3 common,

\(a^3 + \frac{1}{a^3 } + 3(a + \frac{1}{a})\)

after substituting the value of \(a + \frac{1}{a }\), the actual equation becomes:

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

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


Option A is the Answer
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Re: If (a + 1/a)^2 = 3, find the value of a^3 + 1/a^3 [#permalink] New post   09 Dec 2019, 18:55
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Bunuel wrote:
If \((a +\frac{1}{a})^2=3\), find the value of \(a^3 + \frac{1}{a^3}\)


A. 0

B. 1

C. \(\sqrt{3}\)

D. \(2+\sqrt{3}\)

E. Not enough information

Are You Up For the Challenge: 700 Level Questions


Since (a + 1/a)^2 = 3, a + 1/a = ±√3. Furthermore, since (a + 1/a)^2 = a^2 + 2a(1/a) + 1/a^2 = a^2 + 2 + 1/a^2 = 3, we see that a^2 + 1/a^2 = 1.

Now, if we multiply a + 1/a and a^2 + 1/a^2 (and assume that a + 1/a = √3), we have:

(a + 1/a)(a^2 + 1/a^2) = √3 x 1

a^3 + 1/a + a + 1/a^3 = √3

a^3 + √3 + 1/a^3 = √3

a^3 + 1/a^3 = 0

If a + 1/a = -√3, we have:

(a + 1/a)(a^2 + 1/a^2) = -√3 x 1

a^3 + 1/a + a + 1/a^3 = -√3

a^3 - √3 + 1/a^3 = -√3

a^3 + 1/a^3 = 0

We see that either way, a^3 + 1/a^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] New post   25 Feb 2022, 05:38
Since :

\(\left(a+\frac{1}{a}\right)^2\ =\ 3,\ \left(a+\frac{1}{a}\right)\ =\ \left(\sqrt{\ 3}or\ -\sqrt{\ 3}\right)\)
Considering \(\left(a+\frac{1}{a}\right)\ =\ \left(\sqrt{\ 3}\right)\)
\(\left(a+\frac{1}{a}\right)^3\ =\ a^3+\frac{1}{a^3}+3\cdot\left(\cdot a+\frac{1}{a}\right)\)
Hence :
\(3\sqrt{\ 3}\ =\ a^3\ +\ \frac{1}{a^3}+\ 3\sqrt{\ 3}\)
\(\ a^3\ +\ \frac{1}{a^3}=\ 0\)
Similarly considering : \(a+\ \frac{1}{a}=\ -\sqrt{\ 3}\)
\(\left(a+\ \frac{1}{a}\right)^3=\ -3\sqrt{\ 3}\)
=\( \ -3\sqrt{\ 3}=\ a^3+\ \frac{1}{a^3}\ -\ 3\sqrt{\ 3}\)
\(\ a^3+\ \frac{1}{a^3}\ =\ 0\)



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Re: If (a + 1/a)^2 = 3, find the value of a^3 + 1/a^3 [#permalink] New post   25 Feb 2022, 06:24
Bunuel wrote:
If \((a +\frac{1}{a})^2=3\), find the value of \(a^3 + \frac{1}{a^3}\)




Are You Up For the Challenge: 700 Level Questions



(a + 1/a)(a^2 + 1/a^2) = √3 x 1

a^3 + 1/a + a + 1/a^3 = √3

a^3 + √3 + 1/a^3 = √3

a^3 + 1/a^3 = 0

If a + 1/a = -√3, we have:

(a + 1/a)(a^2 + 1/a^2) = -√3 x 1

a^3 + 1/a + a + 1/a^3 = -√3

a^3 - √3 + 1/a^3 = -√3

a^3 + 1/a^3 = 0

We see that either way, a^3 + 1/a^3 = 0

Therefore IMO A
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Re: If (a + 1/a)^2 = 3, find the value of a^3 + 1/a^3 [#permalink] New post   17 Nov 2022, 20:04
Bunuel wrote:
If \((a +\frac{1}{a})^2=3\), find the value of \(a^3 + \frac{1}{a^3}\)


A. 0

B. 1

C. \(\sqrt{3}\)

D. \(2+\sqrt{3}\)

E. Not enough information

Are You Up For the Challenge: 700 Level Questions


(Step 1)

Take the square root of a variable expression squared

|a + (1/a)| = sqrt(3)

a + (1/a) = +sqrt(3) —or— (-)sqrt(3)

(Step 2)

Take the expression (a) + (1/a) and CUBE it

(a + 1/a)^3 = (a)^3 + (1/a)^3 + (3)(a)(1/a) (a + 1/a)


The term: (3)(a)(1/a) = (3)(a/a) = (3)(1) = 3

And you can substitute each possible value of (a + 1/a)
Positive square root of 3
Or
Negative square root of 3

(+sqrt(3))^3 = (a)^3 + 1/a^3 + (3) (+sqrt(3))

Square root of 3 raised to the 3rd power is equal to = (3) (+sqrt(3))

So you end up with

(a)^3 + 1/a^3 = 0

Same result if you put in (-)sqrt(3)

Answer

0

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Re: If (a + 1/a)^2 = 3, find the value of a^3 + 1/a^3 [#permalink] New post   18 Nov 2022, 20:10
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Bunuel wrote:
If \((a +\frac{1}{a})^2=3\), find the value of \(a^3 + \frac{1}{a^3}\)


A. 0

B. 1

C. \(\sqrt{3}\)

D. \(2+\sqrt{3}\)

E. Not enough information


The best way would be to use \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\)

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

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

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

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

A
However, what would be a?
a can not be 0 because 1/a will be undefined.
If a<0, then \(a^3+\frac{1}{a^3}<0\).
If a>0, then \(a^3+\frac{1}{a^3}>0\).
So, only way \(a^3+\frac{1}{a^3}=0\) is when a is some imaginary number and we do not deal with imaginary number in GMAT.
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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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