If (r+1/r)^2 = 5, what is the value of (r^3+1/r^3)^2?

(r + 1/r)² = 5
r + 1/r = √5

(r + 1/r)² = 5
r² + 1/r² + 2(r)(1/r) = 5
r² + 1/r² + 2 = 5
r² + 1/r² = 3

Multiplying the blue equation and the red equation, we get:
(r² + 1/r²)(r + 1/r) = 3√5
r³ + (r²)(1/r) + (1/r²)(r) + 1/r³ = 3√5
r³ + (r + 1/r) + 1/r³ = 3√5

Substituting r + 1/r = √5 into the green equation above, we get:
r³ + √5 + 1/r³ = 3√5
r³ + 1/r³ = 2√5
(r³ + 1/r³)² = (2√5)²
(r³ + 1/r³)² = 20

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\({\left( {r + {1 \over r}} \right)^2} = 5\,\,\,\,\,\left( * \right)\)

\(? = \,{\left( {{r^3} + {1 \over {{r^3}}}} \right)^2}\)

\(?\,\,\, = \,\,\,{\left[ {\left( {r + {1 \over r}} \right)\left( {{r^2} - r \cdot {1 \over r} + {1 \over {{r^2}}}} \right)} \right]^{\,2}} = {\left( {r + {1 \over r}} \right)^2}{\left( {{r^2} - 1 + {1 \over {{r^2}}}} \right)^2}\,\,\,\mathop = \limits^{\left( * \right)} \,\,\,5 \cdot {\left( {{r^2} - 1 + {1 \over {{r^2}}}} \right)^2}\,\,\,\mathop = \limits^{\left( {**} \right)} \,\,\,20\)

\(\left( {**} \right)\,\,{r^2} - 1 + {1 \over {{r^2}}} = \left( {{r^2} + 2 + {1 \over {{r^2}}}} \right) - 3 = {\left( {r + {1 \over r}} \right)^2} - 3\,\,\,\mathop = \limits^{\left( * \right)} \,\,\,2\)


The correct answer is (E).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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I would go for E.
Please show a quick way.

Posted from my mobile device

It is pretty straight forward if dealt this way -

(r+1/r)^3 = r^3 +(1/r)^3 + 3(r*(1/r))(r+(1/r))
(r+1/r)^3 = r^3 +(1/r)^3 + 3(r+(1/r))

We know that r+(1/r) = (5)^1/2

So r^3 +(1/r)^3 = 2*(5^(1/2))
Square the ans = 20

Hi GMATGuruNY

should not the highlighted part above be |r + 1/r| = √5

Why did you avoid modulus sign?

Since \(r + \frac{1}{r} = √5\) yields one of the answer choices, there is no need to consider \(r + \frac{1}{r} = -√5\).
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