Given:
[tex]x^{10}-2x^5+1=0[/tex]
Step 1: To identify the quadratic form of the given equation.
[tex]\begin{gathered} x^{10}-2x^5+1=0 \\ (x^5)^2-2x^5+1=0 \\ \text{Put x}^5=t,\text{ it gives} \\ t^2-2t+1=0 \end{gathered}[/tex]
So, t = x²
Step 2: Factor the quadratic equation in step 1.
[tex]\begin{gathered} t^2-2t+1=0 \\ t^2-t-t+1=0 \\ t(t-1)-t(t-1)=0 \\ (t-1)(t-1)=0 \end{gathered}[/tex]
Thus, the factors of the equation is
[tex](t-1)(t-1)=0[/tex]
Step3: solve for x.
[tex]\begin{gathered} (t-1)(t-1)=0 \\ (x^5-1)(x^5-1)=0 \\ \Rightarrow x^5-1=0,x^5-1=0 \\ \Rightarrow x=1 \end{gathered}[/tex]
Answer: x = 1