help algebra

Given that x satisfies $x^2 - 5x + 1 = 0$,
find the value of $x^4 +\dfrac{1}{ x^4 }$.

$\begin{array}{|rcll|} \hline \mathbf{x^2 - 5x + 1} &=& \mathbf{0} \\\\ x_{1,2} &=& \dfrac{5\pm\sqrt{25-4}}{2} \\ x_{1,2} &=& \dfrac{5\pm\sqrt{21}}{2} \\ \hline x_{1,2}^2 &=& \left( \dfrac{5\pm\sqrt{21}}{2} \right)^2 \\ x_{1,2}^2 &=& \dfrac{123\pm 5\sqrt{21}}{2}\\ \hline x_{1,2}^4 &=& \left( \dfrac{123\pm 5\sqrt{21}}{2} \right)^2 \\ x_{1,2}^4 &=& \dfrac{527\pm 115\sqrt{21}}{2} \\ \hline \end{array}$

$x_1:$

$\begin{array}{|rcll|} \hline x_1^4 + \dfrac{1}{x_1^4} &=& \dfrac{527+115\sqrt{21}}{2} +\dfrac{2}{527+115\sqrt{21}} \\\\ x_1^4 + \dfrac{1}{x_1^4} &=& \dfrac{527+115\sqrt{21}}{2} +\dfrac{2}{(527+115\sqrt{21})}*\dfrac{(527-115\sqrt{21})}{(527-115\sqrt{21})} \\\\ x_1^4 + \dfrac{1}{x_1^4} &=& \dfrac{527+115\sqrt{21}}{2} +\dfrac{2(527-115\sqrt{21})}{4} \\\\ x_1^4 + \dfrac{1}{x_1^4} &=& \dfrac{527+115\sqrt{21}}{2} +\dfrac{(527-115\sqrt{21})}{2} \\\\ x_1^4 + \dfrac{1}{x_1^4} &=& \dfrac{527+115\sqrt{21}+527-115\sqrt{21}}{2} \\\\ x_1^4 + \dfrac{1}{x_1^4} &=& \dfrac{2*527}{2} \\\\ x_1^4 + \dfrac{1}{x_1^4} &=& 527 \\ \hline \end{array}$

$x_2:$

$\begin{array}{|rcll|} \hline x_2^4 + \dfrac{1}{x_2^4} &=& \dfrac{527-115\sqrt{21}}{2} +\dfrac{2}{527-115\sqrt{21}} \\\\ x_2^4 + \dfrac{1}{x_2^4} &=& \dfrac{527-115\sqrt{21}}{2} +\dfrac{2}{(527-115\sqrt{21})}*\dfrac{(527+115\sqrt{21})}{(527+115\sqrt{21})} \\\\ x_2^4 + \dfrac{1}{x_2^4} &=& \dfrac{527-115\sqrt{21}}{2} +\dfrac{2(527+115\sqrt{21})}{4} \\\\ x_2^4 + \dfrac{1}{x_2^4} &=& \dfrac{527-115\sqrt{21}}{2} +\dfrac{(527+115\sqrt{21})}{2} \\\\ x_2^4 + \dfrac{1}{x_2^4} &=& \dfrac{527-115\sqrt{21}+527+115\sqrt{21}}{2} \\\\ x_2^4 + \dfrac{1}{x_2^4} &=& \dfrac{2*527}{2} \\\\ x_2^4 + \dfrac{1}{x_2^4} &=& 527 \\ \hline \end{array}$

$x^4 +\dfrac{1}{ x^4 } =527$