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

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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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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.

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

0

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

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

$$(a + \frac{1}{a})(a^2 + \frac{1}{a^2}) = \sqrt{3}$$
$$a^3 + a + \frac{1}{a} + \frac{1}{a^3} = \sqrt{3}$$
$$a^3 + \frac{1}{a^3} + \sqrt{3} = \sqrt{3}$$
$$a^3 + \frac{1}{a^3} = 0$$­
If (a + 1/a)^2 = 3, find the value of a^3 + 1/a^3 [#permalink]
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