Q.Solve the equation $x^5 - 5x^4 + 9x^3 - 9x^2 + 5x - 1 = 0$.

Spot $x=1$ as an obvious root, reduce to a palindromic quartic, then solve it with the substitution $t=x+\dfrac1x$.

$x^5-5x^4+9x^3-9x^2+5x-1=0.$

Step 1 — find an obvious root. Testing $x=1$: $1-5+9-9+5-1=0$. So $x=1$ is a root.

Step 2 — divide by $(x-1)$. Synthetic division of the coefficients $1,-5,9,-9,5,-1$ by root $1$ gives quotient coefficients $1,-4,5,-4,1$ (remainder $0$), i.e.

$x^5-5x^4+9x^3-9x^2+5x-1 = (x-1)\left(x^4-4x^3+5x^2-4x+1\right).$

Step 3 — solve the quartic. $x^4-4x^3+5x^2-4x+1=0$ has palindromic (symmetric) coefficients $1,-4,5,-4,1$, so $x=0$ is not a root and we may divide through by $x^2$:

$x^2-4x+5-\frac4x+\frac1{x^2}=0 \ \Rightarrow\ \left(x^2+\frac1{x^2}\right)-4\left(x+\frac1x\right)+5=0.$

Let $t=x+\dfrac1x$, so $x^2+\dfrac1{x^2}=t^2-2$:

$(t^2-2)-4t+5=0 \ \Rightarrow\ t^2-4t+3=0 \ \Rightarrow\ (t-1)(t-3)=0 \ \Rightarrow\ t=1\text{ or }t=3.$

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