Yes, you could always try that ....

Algebra Level 3

If $x$ is a real number satisfying the equation

$x^{3} + \dfrac{1}{x^{3}} = 2\sqrt{5}$ ,

then what is the value of

$x^{16} + \dfrac{1}{x^{16}}$ ?

The answer is 2207.

1 solution

Brian Charlesworth
Sep 2, 2014

First, we have that

$(x^{3} + \dfrac{1}{x^{3}})^{2} = (2\sqrt{5})^{2} \Longrightarrow x^{6} + 2 + \dfrac{1}{x^6} = 20 \Longrightarrow x^{6} + \dfrac{1}{x^{6}} = 18$ .

Next, note that $(x^2 + \dfrac{1}{x^{2}})^{3} - 3(x^{2} + \dfrac{1}{x^{2}}) = x^{6} + \dfrac{1}{x^{6}}$ . So, letting $y = x^{2} + \dfrac{1}{x^{2}}$ , we have that

$y^{3} - 3y = 18$ , for which $y = 3$ is the only real solution.

Next, we have that $(x^{2} + \dfrac{1}{x^{2}})^{2} = 9 \Longrightarrow x^{4} + \dfrac{1}{x^{4}} = 7$ .

Then $(x^{4} + \dfrac{1}{x^{4}})^2 = 49 \Longrightarrow x^{8} + \dfrac{1}{x^{8}} = 47$ ,

$(x^{8} + \dfrac{1}{x^{8}})^{2} = 2209 \Longrightarrow x^{16} + \dfrac{1}{x^{16}} = \boxed{2207}$ .

Since $x^3 = \sqrt{5} \pm 2$ , we get $x = \frac{\sqrt{5} \pm 1}2$ . Perhaps you started with the golden ratio?

Patrick Corn - 6 years, 9 months ago

Perhaps. :) That's one of the reasons why I chose the title, (the other being that one could 'cheat' and use WolframAlpha, etc.). I just wanted to write up a solution that didn't involve solving for $x$ directly, thus testing one's algebra skills.

Brian Charlesworth - 6 years, 9 months ago

How you derived the second step in your solution ?

Tushar Malik - 6 years, 3 months ago

@Tushar Malik – Note that $(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)$ , and so

$(a + b)^{3} - 3ab(a + b) = a^{3} + b^{3}.$

Now let $a = x^{2}$ and $b = \dfrac{1}{x^{2}}$ and note that we then have $ab = 1.$

Brian Charlesworth - 6 years, 3 months ago

I love your solution.

Panya Chunnanonda - 6 years, 7 months ago

Yes, you could always try that ....

The answer is 2207.

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