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$\begin{aligned} x + \frac 1x & = 3 & \small \color{#3D99F6} \text{Squaring both sides} \\ x^2 + 2 + \frac 1{x^2} & = 9 \\ x^2 + \frac 1{x^2} & = 7 & \small \color{#3D99F6} \text{Squaring both sides} \\ x^4 + 2 + \frac 1{x^4} & = 49 \\ x^2 + \frac 1{x^2} & = 47 \\ \implies z & = \boxed{47} \end{aligned}$

$\text{}x+$ $\frac{1}{x}=a$ $\implies$ $\text{}x^{4}+$ $\frac{1}{x^{4}}={a^{4}}-{4{a^{2}}}+2$

we have :

$\text{}x+$ $\frac{1}{x}=3$ $\implies$ $\text{}x^{4}+$ $\frac{1}{x^{4}}=z$

Solve for $z$ .

For $\text{}x+$ $\frac{1}{x}=3$ $\implies$ $\text{}x^{4}+$ $\frac{1}{x^{4}}= {3^{4}}-{4×{3^{2}}}+2={\boxed{47}}$

$x+\dfrac{1}{x}=3$

squaring both sides, we have

$\left(x+\dfrac{1}{x}\right)^2=3^2$ $\color{#D61F06}\implies$ $x^2+2(x)\left(\dfrac{1}{x}\right)+\left(\dfrac{1}{x}\right)^2=9$ $\color{#D61F06}\implies$ $x^2+2+\dfrac{1}{x^2}=9$ $\color{#D61F06}\implies$ $x^2+\dfrac{1}{x^2}=7$

squaring both sides, we have

$\left(x^2+\dfrac{1}{x^2}\right)^2=7^2$ $\color{#D61F06}\implies$ $x^4+2(x^2)\left(\dfrac{1}{x^2}\right)^2+\left(\dfrac{1}{x^2}\right)^2=49$ $\color{#D61F06}\implies$ $x^4+\dfrac{1}{x^4}=49-2=\color{#69047E}\large{\boxed{47}}$