Insufficient 5th degree!

$\text{i am calculating the answer using approximation. Let }$

$\begin{aligned} f(x) & = x^5 -x^3+x-2 \\ f^{'}(x) & = 5x^4-3x^2+1 \end{aligned}$

$\boxed {f^{'}(x) = \displaystyle \lim_{\Delta x \to 0} \dfrac{f(x+\Delta x) - f(x)}{\Delta x } \\ f(x+\Delta x) \approx f(x) + \Delta x f^{'}(x) }$

$\text{above } f(2) \text{function increases rapidly } \\ \text{below } f(-1) \text{function decreases rapidly }$

$\begin{aligned} f(2) & = 32-8 =24 \quad \small\color{#3D99F6}{\text{a root between 1 and 2}} \\ f(1) & = -1 \quad \small\color{#3D99F6}{\text{We can safely conclude that }} \lfloor \alpha \rfloor = 1 \\ f(0) & = -2 \\ f(-1) & = -3 \\ \end{aligned}$

$\begin{aligned} f(1+\Delta x) & = f(1)+\Delta x f^{'}(x)= 0 \\ \Delta x & = -\dfrac{f(1)}{f^{'}(1)} = \dfrac{1}{3} \\ f(1+\Delta x+\Delta x_2 ) & = f(1+\Delta x )+\Delta x_2\times f^{'}(1+\Delta x)= 0 \quad \small\color{#3D99F6}{\text{Repeating the process 2nd time gives better accuracy }} \\ \Delta x_2 & = -\dfrac{f(1+\Delta x)}{f^{'}(1+\Delta x)} = -\dfrac{f\left(\frac43\right)}{f^{'}\left(\frac43\right)} = -0.1026 \end{aligned}$

$\begin{aligned} \alpha & = 1+\Delta x+\Delta x_2 \\ & = 1+\dfrac{1}{3}-0.1026 = 1.2307 \\ \alpha^{6} & = 3.031 \\ \lfloor \alpha^6 \rfloor & = 3 \end{aligned}$