If x - (1/x) = sqrt{13},then what is the valu | vedclass

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Question

(C) Given $x - (1/x) = \sqrt{13}$.
First,find $x^2 + (1/x^2)$:
$(x - 1/x)^2 = x^2 + 1/x^2 - 2 = (\sqrt{13})^2 = 13$.
So,$x^2 + 1/x^2 = 13 + 2 = 15$

Vedclass · Accepted Answer

$239\sqrt{13}$

Vedclass · Answer

(C) Given $x - (1/x) = \sqrt{13}$.
First,find $x^2 + (1/x^2)$:
$(x - 1/x)^2 = x^2 + 1/x^2 - 2 = (\sqrt{13})^2 = 13$.
So,$x^2 + 1/x^2 = 13 + 2 = 15$.
Next,find $x^2 + (1/x^2)$ is not enough,we need $x^2 + 1/x^2$ and $x^3 - 1/x^3$.
We know $(x - 1/x)^3 = x^3 - 1/x^3 - 3(x - 1/x)$.
$(\sqrt{13})^3 = x^3 - 1/x^3 - 3(\sqrt{13})$.
$13\sqrt{13} = x^3 - 1/x^3 - 3\sqrt{13} \implies x^3 - 1/x^3 = 16\sqrt{13}$.
Now,$(x^2 + 1/x^2)(x^3 - 1/x^3) = x^5 - 1/x + x - 1/x^5 = (x^5 - 1/x^5) + (x - 1/x)$.
$(15)(16\sqrt{13}) = (x^5 - 1/x^5) + \sqrt{13}$.
$240\sqrt{13} = (x^5 - 1/x^5) + \sqrt{13}$.
$x^5 - 1/x^5 = 240\sqrt{13} - \sqrt{13} = 239\sqrt{13}$.

If $x - (1/x) = \sqrt{13}$,then what is the value of $x^5 - (1/x^5)$?

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