Q.Solve the equation: $6x^6 - 25x^5 + 31x^4 - 31x^2 + 25x - 6 = 0$.

The coefficients are anti-symmetric (an "odd reciprocal equation"), so $x=1$ and $x=-1$ are roots; factor those out, then reduce the remaining palindromic quartic using the substitution $y=x+\tfrac1x$.

Write $f(x)=6x^6-25x^5+31x^4+0\cdot x^3-31x^2+25x-6$. Notice the coefficients $6,-25,31,0,-31,25,-6$ satisfy $a_k=-a_{6-k}$ — an odd (anti-palindromic) reciprocal equation, for which $x=1$ and $x=-1$ are always roots.

Check: $f(1)=6-25+31+0-31+25-6=0$ ✓, and $f(-1)=6+25+31+0-31-25-6=0$ ✓.

Divide out $(x-1)(x+1)=x^2-1$: long division gives

$f(x) = (x^2-1)(6x^4-25x^3+37x^2-25x+6)$

Solve the quartic $6x^4-25x^3+37x^2-25x+6=0$, which is palindromic (coefficients $6,-25,37,-25,6$ read the same forwards/backwards). Divide through by $x^2$:

$6\left(x^2+\dfrac{1}{x^2}\right) - 25\left(x+\dfrac{1}{x}\right) + 37 = 0$

Let $y=x+\dfrac1x$, so $x^2+\dfrac{1}{x^2}=y^2-2$:

$6(y^2-2)-25y+37=0 \;\Rightarrow\; 6y^2-25y+25=0$

Solving: $y = \dfrac{25\pm\sqrt{625-600}}{12} = \dfrac{25\pm5}{12}$, giving $y=\dfrac{5}{2}$ or $y=\dfrac{5}{3}$.

Case $y=\dfrac{5}{2}$: $x+\dfrac1x=\dfrac52 \Rightarrow 2x^2-5x+2=0 \Rightarrow x=\dfrac{5\pm3}{4}$, giving $x=2$ or $x=\dfrac12$.

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