An algebra problem by Mohammad Khaza

$\begin{aligned} x - \frac 1x & = 4 \\ \left(x - \frac 1x \right)^2 & = 4^2 \\ x^2 - 2 + \frac 1{x^2} & = 16 \\ x^2 + \frac 1{x^2} & = 18 \\ \left(x^2 + \frac 1{x^2}\right)^2 & = 18^2 \\ x^4 + 2 + \frac 1{x^4} & = 18^2 \\ x^4 + \frac 1{x^4} & = \boxed{322} \end{aligned}$

Working on $x -$ $\dfrac{1}{x}$ $= 4$

$x -$ $\dfrac{1}{x}$ $= 4$

$\Rightarrow$ $(x -$ $\dfrac{1}{x})^2$ $= 4^2$

Applying the formula: $(a - b)^2 = a^2 - 2ab + b^2$

$\Rightarrow x^2 - 2$ $x \cdot$ $\dfrac{1}{x}$ $+$ $(\dfrac{1}{x})^2$ $= 16$

$\Rightarrow x^2 - 2 +$ $\dfrac{1}{x^2}$ $= 16$ $\Rightarrow x^2 +$ $\dfrac{1}{x^2}$ $= 18$

$\Rightarrow (x^2 +$ $\dfrac{1}{x^2})^2$ $= 18^2$

$\Rightarrow x^4 + 2 +$ $\dfrac{1}{x^4}$ $= 18^2$

Therefore, $x^4 +$ $\dfrac{1}{x^4}$ $= 18 \times 18 - 2 = \boxed{322}$

$\displaystyle x - \frac { 1 } { x } = 4 \rightarrow x ^ { 2 } - 1 = 4x \rightarrow x ^ { 2 } -4x - 1 = 0 \rightarrow x = \frac { 4 \pm \sqrt { -4 ^ { 2 } - 4 \times 1 \times -1 } } { 2 \times 1 } \rightarrow x = \frac { 4 \pm \sqrt { 16 + 4 } } { 2 } \rightarrow x = 2 \pm \sqrt { 5 } \rightarrow ( 2 \pm \sqrt { 5 } ) ^ { 4 } + \frac { 1 } { ( 2 \pm \sqrt { 5 } ) ^ { 4 } } \rightarrow \boxed { 322 }$