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

Question

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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Ansh777 · Accepted Answer

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

RajKomanapalli · Answer

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

ScottTargetTestPrep · Answer

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

GmatPoint · Answer

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

﻿

Crytiocanalyst · Answer

(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

Fdambro294 · Answer

(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

Posted from my mobile device

chetan2u · Answer

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.

Regor60 · Answer

Square roots. Ewww.

Expand left side:

a^2+2+1/a^2 = 3 and:

a^2+1/a^2 =1

Now:

(a^2+1/a^2)(a+1/a) =

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

Substitute from above:

(1)(a+1/a) =

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

Or

a^3+1/a^3 = 0

Posted from my mobile device

ZIX · Answer

Here's an easier approach:

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

With this we get two equations:

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

Multiply eq. 1 and 2

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

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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

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